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Cornell UntDetsitg iCibtatg 1 JItltaca, Ntw f orb BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF HENRY W. SAGE 1891 arV19569 Elementary calculus Cornell University Library 3 1924 031 220 811 The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031220811 ELEMENTARY CALCULUS THE MACMILLAN COMPANY NEW YORK • BOSTON • CHICAGO • DALLAS ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY ■ CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO ELEMENTARY CALCULUS BY WILLIAM F. OSGOOD, Ph.D., LL.D PEKKIN8 PK0FBS80K OF MATHEMATICS IN HABVAKD UNIVERSITY Wetogorft THE MACMILLAN COMPANY 1921 AU HgMe reeemeel 5 COPTBIGHT, 1921, bt the macmillan company. Set up and electrotyped. Published January, 1931. J. S. Gushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE The object of this book is to present the elements of the Differential Calculus in a form easily accessible for the under- graduate. It is possible, from the very beginning, to illustrate the ideas and methods of the Calculus by means of applications to physics and geometry, which the student can readily grasp, and which will seem to him of interest and value. To do this, the stress in the illustrative examples worked in the text must be laid first of all on the thought which underlies the method of solution, in distinction from the exposition of a process, re- duced in the worst teaching to rules, whereby the answer can be obtained. The 'treatment of maxima and minima. Chapter III, §§ 2, 3, and curve tracing, Chapter III, § 5 and Chapter VII, § 10, will serve to show what is here meant. It is, however, also essential that the student receive thorough training in the formal processes and the technique of the Cal- culus, and this side has been treated with care and complete- ness. Note, for example, the differentiation of composite functions in Chapter II, § 8, and the exposition of the use of differentials in differentiating in Chapter IV, §§ 4, 5. An important application of the graphical inethods, with which the Calculus is so intimately associated, is that of solving approximately numerical equations which do not come under the standard rules of algebra and trigonometry. Hitherto, however, little attempt has been made to present this subject, simple as it is, in any systematic and elementary manner. In Chapter VII the common methods in use by physicists and others who apply the Calculus are set forth and illustrated by simple examples. The book might have included a brief treatment of curva- ture and evolutes, and the cycloid. But probably most VI PREFACE teachers of the Calculus will prefer to take up integration next, and so the closing chapter is devoted to the last of the elementary functions, the inverse trigonometric functions, with special reference to their one great application in the elements of mathematics, namely, their application to integration. The book is so written that it can be adapted, if desired, to an abridged course, in which, after the fundamentals of the first three chapters have been covered, any of the remaining topics can be treated briefly, and thus a wide scope in subject matter is possible, even when the time is short. Cambridge, Massacbusbtts, January, 1921. CONTENTS CHAPTER I INTRODUCTION 1. Functions 2. Continuation. General Definition of a Function PAGE 1 10 CHAPTER II DIFFERENTIATION OF ALGEBRAIC FUNCTIONS. GENERAL THEOREMS 1. Definition of the Derivative . 2. Differentiation oi x" . 3. Differentiation of a Constant 4. Differentiation ot '^x . 5. Three Theorems about Limits. Infinity 6. General Formulas of Differentiation 7. General Formulas of Differentiation, Continued 8. General Formulas of Differentiation, Concluded 9. Differentiation of Implicit Algebraic Functions 13 16 20 21 22 29 32 35 39 CHAPTER III APPLICATIONS 1. Tangents and Normals .46 2. Maxima and Minima 49 3. Continuation : Auxiliary Variables 53 4. Increasing and Decreasing Functions 60 5. Curve Tracing ... 64 6. Relative Maxima and Minima. Points of Inflection . . 67 7. Necessary and Sufficient Conditions 71 8. Velocity; Rates 72 vii VUl CONTENTS CHAPTER IV INFINITESIMALS AND DIFFERENTIALS PAGE 1. Infinitesimals ... 81 2. Continuation. Fundamental Theorem .... 87 3. Differentials 91 4. Technique of Differentiation 95 5. Continuation. Differentiation of Composite Functions . 100 CHAPTER V TRIGONOMETRIC FUNCTIONS 1. Radian Measure .... .... 105 2. Differentiation of sin x 110 3. Certain Limits 112 4. Critique of the Foregoing Differentiation .... 116 5. Differentiation of cos x, tan x, etc 117 6. Shop Work . 118 7. Maxima and Minima 121 8. Tangents in Polar Coordinates 128 9. Differential of Arc 134 10. Rates and Velocities 138 CHAPTER VI LOGARITHMS AND EXPONENTIALS 1. Logarithms 146 2. Differentiation of Logarithms 151 1 3. The Limit hm (1+0' 155 4. The Compound Interest Law 156 5. Differentiation of e^ 167 6. Graph of the Function x" 160 7. The Formulas of Differentiation to Date .... 161 CHAPTER VII APPLICATIONS 1. The Problem of Niunerical Computation .... 166 2. Solution of Equations. Known Graphs .... 166 CONTENTS IX PAGE 3. Interpolation 170 4. Newton's Method ... .... 172 5. Direct Use of the Tables 176 6. Successive Approximations 180 7. Arrangement of the Numerical Work in Tabular Form . 184 8. Algebraic Equations 187 9. Continuation. Cubics and Biquadratics . . 189 10. Curve Plotting . . 195 CHAPTER VIII THE INVERSE TRIGONOMETRIC FUNCTIONS 1. Inverse Functions ... ... 206 2. The Inverse Trigonometric Functions . . • 209 3. Shop Work 215 4. Continuation. Numerical Computation .... 218 5. Applications 221 CALCULUS CHAPTER I INTRODUCTION The Calexilus was invented in the seventeenth century by the mathematician, astronomer, and physicist, Sir Isaac Newton in England, and the philosopher Leibniz in Germany. The reaction of the invention on geometry and mathematical physics was most important. In fact, by far the greatest part of the mathematics and the physics of the present day owes its existence to this invention. 1. Functions. The word function, in mathematics, was first applied to an expression involving one or more letters which represent variable quantities ; as, for example, the expressions (a) a?, 23?-Z'x + l; (6) V«, Va'^ — x^ ; ,. x^ xy ax + by . a + x x^ + y^ Va;2 ^yij^e2 (d) sin a;, logs;, tan^isc. In the second example under (6), two letters enter ; but a is thought of as chosen in advance and then held fast, x alone being variable. A quantity of this kind is called a constant. Thus ax + o is a function of x which depends on two constants, a and 6. 1 2 ' CALCULUS Such expressions are written in symbolic, or abbreviated, form as f(x), f(x, y) (read : "/ of x," "f of x and y " etc.) ; other letters in common use being F, <f>, $, etc.* Thus the equation (1) f{x) = 2a?-'&x + l defines the function /(«) in the present case to be 2a!'— 3a;+l. Again, (2) <j,{x, y, e) = x"- + y' -\- «• is an equation defining the function <j>{x, y, z) as x^ + y^ + z*. We shall be concerned for the present with functions of one single variable, as illustrated by (1) above. Here, x is called the independent variable, since we assign to it any value we like. The value of the function, or more briefly, the function, is called the dependent variable, and is often denoted by a single letter, as ^ =/(«) or y = 23? — Zx + l. Graphs. A function of a single variable, y =/(<»), can be represented geometrically by its graph, and this repre- sentation is of great aid in studying the proper- ties of the function. The independent vari- able is laid off as the as-coordinate, or ab- scissa, and the depend- _ ^ ent variable, or func- * To distinguish between /(x) and F(x), read the first " small/ of x " and the second, " Uu^e Fotx." INTRODUCTION 3 tion, as the y^oordinate, or ordinate. Thus the graph of the function , , is the curve Illustrations from Geometry and Physics. The familiar formulas of geometry and physics afford simple examples of functions. Thus the area, A, of a circle is given by the formula A = ■jtr\ where r denotes the radius, n being the fixed number 3.1416. Here, r is thought of as the independent variable, — it may have any positive value whatever, — and A is the function, or dependent variable. Again, for the three round bodies, the volumes are : (a) F= firr^, sphere ; (6) F=irr%, cylinder; (c) V='^r%, cone. In (6) and (c), h denotes the altitude and r, the radius of the base ; V is here a function of the two independent variables, r and h. The surfaces of these bodies are given by the formulas : (a) S = Attt^, sphere ; (/8) S = 2Trrh, cylinder ; (y) S = tttI, cone ; I, in the last formula, denotiag the slant height. Thus we have three further examples of functions of one or of two variables. The formula for a freely falling body is s = \gt\ where s denotes the distance fallen and t the time ; $> is a constant, for it has just one value after the units of time and 4 CALCULUS length liave been chosen. Here, t is the independent variable and 8 is the function. If, however, we solve this equation for f : , then s becomes the radependent variable and t, the function. Sometimes two variables are connected by an equation, as pv = c, where p denotes the pressure of a gas and v its volume, the temperature remaining constant. Here, either variable can be chosen as the independent variable, and when the equation is solved for the other variable, the latter becomes the de- pendent variable, or function. Thus, if we write c P p is the independent variable, and v is expressed as a function of p. But if we write c P = -, V the rSles are reversed. The Independent Variable Bestricted. Often the independent variable is restricted to a certain interval, as in the case of the function y = Vo^ — x\ Here, x must lie between — a and a : ■ a<^ x<^a. since other values of x make a^ — x^ negative, and the above expression has no meaning. This was also the case with the geometric examples above cited. There, r, h, I were necessarily positive, since there is no such thing, for example, as a sphere of zero or negative radius. The independent variable may also be restricted to being a positive whole number, as in the case of the sum of the first n INTRODUCTION terms of a geometric progression : ' s„ = a + ar + ar^ + — + ar'-K Here, a— ar" 1 — r Suppose a = 1, r = ^, the progression thus becoming Then 1 + 1 + 1+.. .+J-. 2 22 ^ 2"-i -i.-^ 2 and we have an example of a function with the iadependent variable a natural number, i.e. a positive integer. In the case of ihe functions treated in the calculus, the do- main of the independent variable is a continuum, i.e., for func- tions of a single variable, an interval, as a-^x<b, or < as. Ordinarily, the later letters of the alphabet, particularly X, y, z, are used to represent variables, the early letters denot- ing constants. Thus it will be understood, when such an ex- pression as aad' + OX + C is written down, that o, 6, c are constants and x is the variable. Multiple-Vahied Functions; Principal Value. The expres- sions above cited are all examples of single-valued functions ; i.e. to each value of the independent variable x corresponds but one value of the function. A function may, however, be muUiple-valued ; as in the case of the function y defined by the equation a,2 + ^2 _ a\ Here y = ± Va^ — x^, 6 CALCULUS and so is a double-valued function. This function is, however, completely represented by means of the two single-valued functions, y = ^a^—x^ and y = — Va* — iC*. Oraphofy,when Graph of f \. x=a Fig. 2 They form the branches of this multiple-valued function. The student should notice that the radical sign y' is defined as meaniQg the positive square root, not either the positive or the negative square root at pleasure. If it is desired to ex- press the negative square root, the minus sign must be written in front of the radical sign, — -y/. Thus Vi = 2, and not — 2. This does not mean that 4 has only one square root. It means that the notation V4 calls for the positive, and not for the negative, of these t;wo roots. Again, and not - 2. For (- And, generally, (1) 2)2= 4, and y/ means the positive root. 1 ■\/s^ = — x, if X is positive ; if X is negative. A similar remark applies to the symbol tJ/, which is like- wise used to mean the positive 2nth root. Moreover, a^=^^, '=VE. The function y = -\/x INTRODUCTION 7 is often called the principal value of the double- valued function defined by the equation Since multiple-valued functions are studied by means of single-valued functions, it -will be understood henceforth, un- less the contrary is explicitly , stated, that the word function means single-valued function. Absolute Value. It is frequently desirable to use merely the numerical, or absolute value of a quantity, and to have a notation for the same. The notation is: |a!|, read " absolute value of X." Thus |-3| = 3 and |3|=3. We can now write in a single formula what was formerly stated by the two equations (1), namely the definition of the radical sign, V • (2) V^ = |a|. Again, by the difference of two numbers we often mean the value of the larger less the smaller. Thus the difference of 4 and 10 is 6 ; and the difference of 10 and 4 is also 6. The difference of a and 6, in this sense, can be expressed as either Continuous Functions. A function, /(«), is said to be con- tinuous if a slight change in x produces but a slight change in the value of the function. Thus the polynomials are readUy shown to be continuous ; cf. Chap. II, § 5, and all the func- tions with which we shall have to deal are continuous, save at exceptional points. As an example of a function which is discontinuous at a certain point may be cited the function (see Tig. 3) /(.)=i. 8 CALCULUS y= When X approaches the value 0, the function increases nu- merically without limit. The graph of the function has the axis of y as an asymptote. The fractional rational functions are continuous except at the points at which the denominator vanishes. 1^ Thus the function ^'-'-^ is continuous except at the points a; = 1 and a; = — 1. Here, the function becomes infinite. Its graph is the curve _ x'^ + 1 y~{x-l)Qc + iy which evidently has the lines x = l and a = — 1 as asymptotes. The function Fig. 3 /(a;) = tan a; is continuous except when x is an odd midtiple of ir/2, ^ 2w + l X ^ -r — TT. EXERCISES 1. If f{x) =x^-4:X + 3, show that /(1) = 0, /(2)=-l, /(3) = 0. Compute /(0),/(4). Plot the graph of the function. 2. If " ^(a;) = 4a!' compute ^(2) and ^(V3). 3. If F(x): 2,x-Z x + 7 ' compute F{-\/2) correct to three significant figures. Ans. -.0204. INTRODUCTION 9 4. If $(«) = (a;' — a;)siii x, find all the values of x for -whicli *(0)=0. 5. If ^(a!)=a!^ — a;"^, findi/-(8). 6. Solve the equation 3^ — xy -\- 3 = By for y, thus expressing ^ as a function of x. 7. If f(x)=ci^, ■ show tlmt /(a')/(2/) =/(» + y)- 8. If y= ^ + ^, express a; as a function of y. 9. Draw the graph of the function /(a;)=cB2 + 4a! + 3, taking 1 em. as the unit. Suggestion : Write the function in the form, (x + l)(a; + 3). 10. Draw the graph of the function f[x) = a^ — 4 a;. 11. Draw the graph of the function and hence illustrate the two discontinuities which this func- tion has. 12. Draw the graph of the function f( X 1 1 ^^^' a? {x-lf 10 CALCULUS 13. For what values of x are the following functions dis- continuous ? (a) /(») = cot » ; (c) f(x) = esc a; ; (6) f(x) = sec a; ; (d) /(«) = tan |- 14. Express the double-valued function defined by the equation a?-f = -l in terms of two single-valued functions. 15. Express the quadruple-valued function defined by the equation y* - 2j/^ + x'' = in terms of four single-valued functions. 16. Express the sum s„ of the first n terms of the arithmetic progression a+{a + b)+(a + 2b)+ — +(a + n — lb) as a function of n. Thus obtain the sum of the first n positive integers as a function of n. 17. If P dollars are put at simple interest for one year at r per cent, (a) express the amount A (principal and interest) as a function of P and r. (b) Express the amount A at the end of n years, the interest being compounded annually, as a function of P, r, and n. (c) Express the amount A at the end of one year, if the interest is compounded m times in the year at equal intervals, as a function of P, r, m. 2. Continuation. General Definition of a Function. The con- ception of the function is broader than that of the mathemati- cal formulas mentioned in the last paragraph. Let us state the definition in its most general form. Definition of a Function. 77ie variable y is said to be a function of the variable x if there eadsts a law whereby, when x is given, y is determined. INTRODUCTION 11 Consider, for example, a quantity of gas confined in a cham- ber, — for instance, the charge of the mixture of gasolene and air as it is being compressed in the cylinder of an automobLle. The charge exerts at each instant a definite pressure, p, of so many pounds per square inch on the walls of the chamber, and this pressure varies with the volume, v, occupied by the charge. In the small fraction of a second under consideration, presumably but little heat is gained or lost through the walls of the chamber, and thus p is a function of v, In this ease, the function is given approximately by the math- ematical formula _ where C denotes a certain constant. But that which is of first importance for our conception is not the formula, but the fact that to each value of v there corresponds a definite value of p. In other words, there is a definite graph of the relation be- tween V and p. The representation of the relation by a math- ematical formula is, indeed, important; but what we must first see clearly is the fact that there is a definite relation to express. As another illustration take the curve traced out by the pen of a self-registering thermometer of the kind used at a meteorological station. The instrument consists of a cylindri- cal drum turned slowly by clock- work at uniform speed about a vertical axis, a sheet of paper ^la. 4 being wound firmly round the drum. A pen is held against the paper, and the height of the pen above a certain level is proportional to the height of the temperature above the temperature corre- 12 CALCULUS sponding to that level. The apparatus is set in operation, and when the drum has been turning for a day, the paper is taken off and spread out flat. Thus we have before us the graph of the temperature for the day ia question, the independent variable being the time (measured in hours from midnight) and the dependent variable being the temperature, represented by the other coordinate of a point on the curve. One more illustration, — that of the resistance of the atmos- phere to a rifle bullet. This resistance, measured in pounds, depends on the velocity of the bullet, and it is a matter of physical experiment to determine the law. But that which is of first importance for our conceptions is the fact that there is a law, whereby, when the velocity, v, is given an arbitrary value withia the limits of the velocities considered, there cor- responds to this V a definite value, B, of the resistance. We say, then, that B is a, function of v and write B = fiv). In this connection, ef. the chapter on Mechanics, § 7, Graph of the Resistance, La the author's Differential and Integral Gdkulus. CHAPTER II DIFFERENTIATION OF ALGEBRAIC FUNCTIONS GENERAL THEOREMS 1. Definition of the Derivative. The Calculus deals with varying quantity. If j/ is a function of x, then x is thought of, not as having one or another special value, but as flowing or growing, just as we think of time or of the expanding cir- cular ripples made by a stone dropped into a placid pond. And y varies with x, sometimes increasing, sometimes decreas- ing. Now if we consider the change in x for a short interval, say from x=:Xotox= x', the corresponding change in y,a,sy goes from y^ to y', will be iu general almost proportional to the change in x. For the ratio of these changes is x' — Xa and this quantity changes only slightly when x' is nearly equal to Xq. Let us study this last statement minutely. Fig. 5 The above ratio has a simple geometric meaning, if we draw the graph of the function ; for PM=x'-Xa; MP' = y'-y^, 13 14 CALCULUS aj'-OTo where t' denotes the angle which the secant PP makes with the axis of x. Now let x' approach ojq as its limit. Then r' approaches as its limit the angle t which the tangent line of the graph at P makes with the axis of x, and hence lim y ~yo = tan r x'^ x' — !Cd ( read : "limit, as x' approaches Xg, of ^ ~^° )• \ X' — Xf, J The determination of this limit and the discussion of its mean- ing is the fundamental problem of the Differential Calculus. Such are the concepts which underlie the idea of the deriva- tive of a function. We turn now to a precise formulation of the definition. Let (1) y=f('») be a given function of x. Let Xq be an arbitrary value of x, and let ^o ^ ^^^ corresponding value of the function : (2) 2/0 =/(ai)). Give to X an Lucrement,* Ax ; i.e. let x have a new value, x', and denote the change in x, namely, a/ — aj^, by Aa; : a;' — a!o = Aa;, a;' = a^ + Aa;. The function, y, will thereby have changed to the value (3) 2/' =/(«'') and hence have received an iucrement. Ay, where y'-yo = Ay, y' = yo + %• • The student must not think of this symbol as meaziing A times x. We might have used a single letter, as h, to represent the difference in question : x'=Xo + h; hut h would not have reminded us that it is the increment of x, and not of y, with which we are concerned. The notation is read " delta x." DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 15 Equation (3) is equivalent to the following : (4) ' yo + Ay =/(a'o + Aaj). From equations (2) and (4) we obtain by subtraction the equation A2/=/(a!o + Aa;)-/(a!o), and hence C5-V Ay^ /(a!o+Aa!)-/(a;o) ^ ^ Ak Aaj Definition oi" a Dbeivativb. The limit which the ratio (5), namely — ^ , approaches when Ax approaches zero : (6) lim^ or li^/^^±Mr^£M, tix^Ax AxiO Ax is called the derivative of y with respect to x and is denoted by Bjy or DJioi) (read : "D x oiy"): (J) ox^Ax In this definition Ax may be negative as well as positive, and the limit (6) must be the same when Aa; approaches from the negative side as when it approaches from the positive side. To differentiate a function is to find its derivative. The geometrical in- terpretation of the analytical process of differentiation is to find the slope of the graph of the function. Por, tanr'=^ y yC^ 1 S^' Ay — ^ ^-ss^^Aa;' vi y X aiQ, i »' Ax Fig. 6 and tanr = lim,tanT' =lim— ^ = Z)^. F'^P ^x^ Ax 16 CALCULUS 2. Differentiation of x". Suppose n has the value 3, so that it is required to diEEerentiate the function (1) y = a?. We must follow the definition of § 1 step by step. Begin, then, by assigning to a; a particxilar value, a^, which is to be held fast during the, rest of the process, and compute from equation (1) the corresponding value yoofy: (2) 2/0 = XgK Next, give to x an arbitrary increment. Ax, denote the corre- sponding increment in y by Ay, and compute it. To this end we first write down the equation (3) ya + Ay={xo+As:y. The right-hand side of this equation can be expanded by the binomial theorem, and hence (3) can be written in a new form :* (4) yo+Ay = 00o^ + 3 x^^Ax + 3 x^Ax^ -(- Aa?. Subtract equation (2) from equation (4) : Ay = Sxg^Ax + 3 XgAx'^ + Aa^. Next, divide through by Ao; : ^ = SV + 3xJ^x + Ax''. Ax We are now ready to let Ax approach as its limit : lim M = lim (3 a;„2 + 3 x^Ax + Ax^). *It is at this point that the specific properties of the functional come into play. Here, it is the'binomial theorem that enables us ultimately to compute the limit. In the differentiations of later paragraphs and chapters it will always h.e some characteristic property of the function in hand which will make possible a transformation at this point. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 17 The limit of the left-hand side is, by definition, D,y. On the right-hand side, each of the last two terms in the parenthesis approaches the limit 0, and so their sum approaches 0, also. The first term does not change with Aw. Hence, the whole parenthesis approaches the limit Swo^. We have, then, as the final result : ^ o a ■0,2/ = 3 V- The subscript has now served its purpose, which was, to remind us that Xq is not to vary with Ax, and it may be dropped. Thus d^^=3xK The differentiation of the function x" in the general case- that n is any positive integer can be carried through in pre- cisely the same manner. As the result of the first step we have (6) ' 2/0 = aso"- Next comes : (6) y„ + Ay = (Xo + Aa;)», and we now apply the binomial theorem to the expression on the right-hand side. Thus (7) yo + ^y = xo" + mV^Ak + '^^^~J'^ a;o»-2Aa;2 H h Aa;». On subtracting (6) from (7) we have : Ay = nXo"-^Ax + ^^"' ~J'^ x^^-^-Ax"- H h Aa;". 1 ■ ^ Now divide through by Ax : ^ = nxo"-^ + '^^"'~'^\ --^Ax +■■■■+ Ax--^ Ax 1 '2 and let Ax approach the limit zero : lim ^ = lim f «a!o"-i + "^" ~ ^^ V'^Aw -H ■ ■ • -I- Aa!»-i\ Arm Ax /^asiO \ 1-2 J Each term of the parenthesis after the first is the product of a constant factor and a positive power of Aa;. This second 18 CALCULUS factor approaches zero when Aa; approaches zero ; consequently the whole term approaches zero. There is only a fixed num- ber of these terms, and so the whole parenthesis approaches the limit waso""". Hence On dropping the subscript we obtain the final result : (8) Z),£B» = raa!»-i. In particular, if n = 1, we have (9) D,x = l. EXERCISES Differentiate the following seven functions, applying the process of § 1 step by step. , 1. y = 4.a?. Ans. Dji = 12x\ 2. y = a^. 3. y = 2f -3a;+.l. Ans. Djy = 4 a; — 3. 4. y = x' — a?. Ans. D^ = 7afi-5!ei. 5.* /(a!)=l-2a^. Ans. DJ(x)=-8a?. 6. ,i>{x)=x-'--2x + l. 7. F{x) = {l-xy. 8. Let y = 5x- x\ and take a;o = 1 ; then yg = 4. If Aa; = .2, then Aj/ = .56 and ^ = 2.8. Show further that, Aa; for Aa;=.l, A;/ = .29, ^=2.9; Aa; and . for Aa; = .01, Ay = .0299, ^=2.99. Aa; * It is immaterial wliethier we write y = l-2x* or f(x) =1-2**. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 19 Plot the curve accurately for values of x from a; = to x = 5, taking 1 cm. as the unit, and draw the secants* in each of the three foregoing cases. What appears to be the slope of the curve at the point {xq, yo) = (l> 4)? Prove your guess to be correct. 9. In Ex. 7, let a!o = — 1- H Ax is given successively the values .01 and — .01, compute Ay and ^• Ax 10. Complete the following table : Ax Ay tanr' = ^ . Ax .1 .01 .001 for each of the functions : (a) y = x^-2x + l, «„ = 2 ; (&) y = x — a?, iCi, = - 1 ; (c) y = 3ce2 — x, cc,, = 0. 11. By means of the general theorem (8) write down the derivatives of the following functions : a^; a^; a!""; x; aP. By means of the definition of § 1 differentiate each of the following functions : Ans. D^= -. 12. 13. 14. 1 y=— X 1 y = Q? Ans. Dji = — Ans. Dji = — * The student shoidd recall from his earlier work how to draw a straight line on squared paper when a point and the slope ot the line are given. 20 CALCULUS 3. Derivative of a Constant. The function where c denotes a constant, has for its graph a right line paral- lel to the axis of x. Since the derivative of a function is repre- sented geometrically by the slope of its graph, it is clear that the derivative of this function is zero : Dji = 0. It is instructive, however, to obtain this result analytically by the process of § 1. We have here : 2/o=/(9i))=c, yo + Ay =/(a!o4- Aa;)= c; hence Aw = and —2 = 0. " Aa; Now allow Aa; to approach 0. The value of Ay/Aa; is always 0, and hence its limit * is : lim^=0, or Z),c = 0. Ax^ Aa; * We note here an error frequently made in presenting the subject of limits in school mathematics. It is there often stated that " a variahle JT approaches a limit Ait X comes indefinitely near to A, but never reaches A." This last requirement is not a part of the conception of a variable's approaching a limit. It is true that it is often inexpedient to allow the independent variable to reach its limit. Thus, in differentiating a func- tion, the ratio Ay/ Ax ceases to have a meaning when As = 0, since divi- sion by is impossible. The problem of differentiation is not to find the value of Ay I Ax, when Ax = ; such a question would be absurd. What we do is to allow Aa; to approach zero as its limit without ever reaching that limit. We can do this for the reason that Az is the iTidepemdent «ario6ie. When, however, it is Ay or Ay /Ax that is under consideration, we have to do with dependevi ■variables, and we have no control over them, as to whether they reach their limit or not. Thus in the case of the text both Ay and Ay/ Ax are constants (=0). When Aa; approaches 0, they always have one and the same value, and so, under the correct concei>tion of approach to a limit each approaches a limit, namely 0. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 21 We can state the result by saying : The derivative of a con- stant is 0. 4. Differentiation of Va;. Let us differentiate y=-\/x. Here, 2/0 = V«0) 2/o + % = Va;o + Aa;, Ay _ Vgp + Aa; — VajQ Ax Aa; We cannot as yet see what limit the right-hand side approaches when Aa; approaches 0, for both numerator and denominator approach 0, and - has no meaning. We can, however, trans- form the fraction by multiplying numerator and denominator by the sum of the radicals and recalling the formula of Elemen- tary Algebra : „= _ 52 =(„ _ 6)(„ + j). Thus % — V^'o + Aa; — Vajp _ Vxg + Aa; -j- Vxp ^a; Aa; Va;o + Aa!+V% _ 1 (xg + Aa;) — a;o 1 ^^ Vxo + AX + Vwo -Vxq + Aa; -|-' V^' and hence lim — ^ = lim — ^ ^ = ^^ • a^ Ax i»^ Va!o -(- Aa; -H Va;o 2 VaJo Dropping the subscript, we have : 2Vm EXERCISES 1. Differentiate the function'^/ = Ans. Dj) = — Vx 2-\/a? 2. If 2/=V2-3a;, 3 show that Z)j« = =^=r ■ 2V2-3a; 22 CALCULUS 3. Prove: D,Vl — x = — 4. Prove: 2)^Va + 6a; = 2Vl-a! 6 2 Va + bx 5. Three Theorems about Limits. Infinity.* In the further treatment of differentiation the following theorems are needed. Theobem I. The limit of the sum of two variables is equal to the sum of their limits: Urn {X+Y)= lim X + lim T. In this theorem we think of X and T as two dependent variables, each of which approaches a limit : limX=^, limr=JB. We do hot care what the independent variable may be. In the applications of the theorem to computing derivatives, the independent variable will always be Ax, and it will be allowed to approach 0, without ever reaching its limit. Since X approaches A, it comes nearer and nearer to this value. Let the difference between the variable and its limit be denoted by £ ; then the limit of e is : (1) X- -A-. = ^, X = = A + €; Similarly, let lime = 0. (2) Y- ■B = =,'?> T= ■■B + r,; then limr , = 0. * This paragraph should he read carefully and its content grasped, but the student should not be required to reproduce it at this stage of his work. He will meet frequent applications of its principles, and he should turn back each time to these pages and read anew the theorem involved, with its proof. When he has thus come to see the full meaning and im- portance of these theorems, he should demand of himself that he be able readily to reproduce the proofs. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 23 It will be convenient to think of these numbers as repre- sented geometrically by points on the scale of numbers, thus : — t — 1 nt — r-t- Fig. 7 Of course, A and B may be negative or 0. e and rj may be negative as well as positive, or even 0. Consider the variable X+Y. Its value from (1) and (2) is : X + T=A + B + € + r,. ^„, , ,. y Hence lim (X+Y) =lim(A + B + e + rj). But since lim e = and lim 17 = 0, the limit of the right-hand ^ side of this equation is A + B, or lim{X + Y)=A + B. Consequently, lim (X+Y)= lim X + lim P, q. e. d. CoBOLLAKY. The limit of the sum of any fined number of variables is equal to the sum, of the limits of these variables : lim (Xi + X2+-+X,)= lim Xi + lim Xj -1- - -1- lim X„. Suppose n = 3. Then Xi + X2 -I- X3 = (Xi -f- X2) -f X3. From Theorem I it follows that lim (Xi -f- X2 + X3) = lim (Xi -|- X2) -1- lim Xs- Applying the Theorem again, we have lim (Xi + Xj) = lim Xj -|- lim Xj. Hence the corollary is true for n = 3. It can how be estab- lished for ji = 4 ; and so on. By the method of Mathematical Induction it can be proven generally. Or, the proof of the main theorem may be extended directly to the present theorem. 24 CALCULUS Theorem II. Tlie limit of the product of two variables is equal to the product of their limits : lim (XF) = (limX)(lim T). From equations (1) and (2) it follows that or XT= AB + B€ + Ari + erj. Hence lim 'XT= lim (AB + B€ + A7i + eq). Since A and B are constants, each of the last three terms in the parenthesis approaches the limit 0, and so the limit of the parenthesis is AB. Hence lim{XY)= AB, or lim {XT) = (lim :X)(lim T), q. e. d. Corollary. The limit of the product of n variables is equal to the product of the limits of these variables : lim (XiXj ... X„) = (lim Xi)(lim X^) - (limX„). The proof is similar to that of the corollary under Theorem I. Remark. As a particular case under Theorem II we have : lim(C'X) = C(limX), where O is a constant. Theobbm III. The limit of the quotient of two variables is equal to the quotient of their limits, provided that the limit of the divisor is not ; ^ , . ^ From equations (1) and (2) above we have : X^ A + t _ Y B + 7, DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 25 Subtract AfB from each side of this equation and reduce : X A_ A-^€ A_ Bt — Ar] Y B B + 7] B B' + Br,' Hence X T ■4- B€ — Ari B' + 5,' and lim^=lim(^- Y \B Be- J B^ + B We wish to show that lim Bc- 1^ = 0. £2 + Br] The numerator is seen at once to approach zero. The limit of the denominator is &. Let fl^ be a positive number less than H B' Fig. 8 B\ Then the denominator will finajly become and remaiu greater than H, and hence the numerical value of the quotient in question will not exceed the numerical value of ■Be- J.T; H But the limit of this expression is zero, and hence ,. X A Y B ,. X limX „ „ ^ or lim — = - — — , q. e. d. Y lim Y In particular, we see that, if a variable approaches unity as its limit, its reciprocal also approaches unity : If limX=l, then lim^ = l. 26 CALCULUS where (7 is a constant and lim X^O. Remark. If the denominator Y approaches as its limit, no general inference about the limit of the fraction can be drawn, as the following examples show. Let T have the values : F=i J- -J- -^ 10' 100' 1000' ' 10"' (1) If the corresponding values of X are : 102' 1002' 10002' ' 10""' X 1 then lim — - = lim — ■ = 0. T 10- 111 (2) If X = Vip' vioo' viooo' ■' ^^' then X/F=10"/^ approaches no limit, but increases beyond all limit. (3) If X = ^, 10' 100' 1000' ' 10"' ' where c is any arbitrarily chosen fixed number, then lim — = c. Y fA\ Tf "V^ J- 1 1 1 *■ '' ~io' ~ioo' iooo' ""10,000' ■"' then X/ Y assumes alternately the values + 1 and — 1, and hence, although remaining finite, approaches no limit. To sum up, then, we see that when X and Y both approach as their limit, their ratio may approach any limit whatever, or it may increase beyond all limit, or finally, although remain- DIFFERENTIATION OP ALGEBRAIC FUNCTIONS 27 ing finite, i.e. always lying between two fixed numbers, no mat- ter how widely tlie latter may differ from each other in value, — it may jump about and so fail to approach a limit. Infinity. If limX=^:?t=0 and limF=0, then X/ Fin- creases beyond all limit, or becomes infinite. A variable Z is said to become infinite when it ultimately becomes and re- mains greater numerically than any preassigned quantity, how- ever large.* If it takes on only positive values, it becomes positively infinite; if only negative values, it becomes negatively infinite. We express its behavior by the notation : limZ = oio or lim^ = -|-oo or limZ = — oo. But this notation does not imply that infinity is a limit ; the variable in this case approaches no limit. And so the notation should not be read " Z approaches infinity" or "Z equals infinity ; but " Z becomes infinite.'' Thus if the graph of a function has its tangent at a certain point parallel to the axis of ordinates, we shall have for that point : , lim — ^ = 00 ; read : " Ay /Ax becomes infinite when Ax approaches 0." Some writers find it convenient to use the expression "a variable approaches a limit " to include the case that the vari- able becomes infinite. We shall not adopt this mode of ex- pression, but shall understand the words " approaches a limit " in their strict sense. '*T!f a function f(x) becomes infinite when x approaches a cer- tain value a, as for example /(a;)=- for a = 0, * Note that the statement sometimes made that " Z becomes greater than any assignable quantity" is absurd. There is no quantity that is greater than any assignable quantity. 28 CALCULUS we denote this by writing /(a)=oo (or /(a) =+00 or =— oo, if this happens to be the case and we wish to call attention to the fact). It is in this sense that the equation tan 90° =00 is to be understood in Trigonometry. The equation does not mean that 90° has a tangent and that the vcdue of the latter is so. It means that, as x approaches 90° as its limit, tana; exceeds numerically any number one may name in advance, and stays above this number as x continues to approach 90° without ever reaching its limit, 90°. Definition of a Continuous Function. We can now make more explicit the definition given iu Chapter I by saying: f(x) is continuous at the poiut a; = a if lim/(a!)=/(a). From Exercises 1-3 below it follows that the polynomials are continuous for all values of x, and that the fractional rational functions are continuous except when the denominator va.nishes. EXERCISES 1. Show that, if n is any positive integer, lim(Z») = (limX)». 2. If O (x) = Co + C^X + C2X^ -\ + cjif, then lim Q{x)= G(a)= (^ + CiU + ofl,^ + - + cjn". x=a 3. If Q{x) and F(x) are any two polynomials and if F{a)^0, then lim^M = ^(^. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 29 4. If X remains finite and Y" approaches as its limit, show ^^"-^ lim(Xr)=0. 5. Show that ^^^ a^ + 1 _ 1 x™3a;2 + 2a;— 1~3" Suggestion. Begin by dividing the numerator and the de- nominator by x^. Evaluate the following limits : 6. lim ^ + W-'^^, 7. Km 12^!+5 ;=»a!3-7a; + 3 ^•^''< .=« 4:afi + 3x*-+7 x^-1 8. lim?^^±^'.„/./. 9. lim^^±^. ^^ x=a cx + da;"' ''' *-«^, x^ ex + dx~^ -^T A- 10. lim ■ \ 11. lim — ' ^ *-" X »=" VS + 5a;2-|- 4»^ 12. lim ^^ + ^' + '^ - 13. lim- X X=o<, VI + a* 6. General Formulas of differentiation. Theorem I. The derivative of the product of a constant and a function is equal to the product of the constant into the deriva- tive of the function : (I) DJ^cu) = cDj.i. Por, let y = cu. Then 2/0 = CMo, yo + Ay = c{ua + Am), hence Ay = cAm, Ay _ Am Aa; Aa;' and lim^ = limfc^\ Aa=M Aa; i»io\_ Ax J 30 CALCULUS The limit of the left-hand side is D^. On the right, Aw/Aa; approaches D^ as its limit. Hence by § 5, Theorem II, the limit of the right-hand side is cD.tt, and we have X>,(cm)=cD,m, q.e.d. Theorem II. Tim derivative of the sum of two functions is equal to the sum of their derivatives : (H) D^{u + v)=D,u + D,v. • For, let y = u + v. Then yo=uo + vo, yo + Ay = Mo + Am + -Wo + Atj, hence Ay = Am + Av, and Ay _ Am Av Ax Ax Ax When Ax approaches 0, the first term on the right approaches Dji and the second D^v. Hence by § 5, Theorem I, the whole right-hand side approaches D^u + D^v, and we have lim^ = lim('^ + ^Vl™ — + lim^, is.,c^Ax Ai=o\Aa; Ax J ^x^Ax ^x^Ax or D^ = D,u + D^v, q. e. d. CoROLLAKT. 2%e derivative of the sum of any number of functions is equal to the sum of their derivatives. If we have the sum of three functions, we can write u + V + w = u -\-(v + w). Hence -D,(m + v+w)= JD^u + Z>,(v + w) = D,u + D,v -t- D^w. Next, we can consider the sum of four functions, and so on. Or we can extend the proof of Theorem II immediately to the sum of n functions. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 31 ■ Polynomials. We are now in a position to differentiate any polynomial. For example: = 7D,3^ - 5Dji? + 1 = 280)3 - 153? + 1. EXERCISES Differentiate the following functions : 1. y = 2x^ — Zx + l. Ans. D^y = ix — 3. 2. y=a + bx+ cx\ Ans. D^ —b + 2cx. 3. y = x^-3<jfi + x — l. Ans. D^ = ia^ — 9 x'- + 1. 4. y = a + bx + CO? + dsfi. 5. y = — — Ans. 3a:^ — 6a!' — 1. ^ 2 6. jr^. ax--^2hx + c ^^^ ax + b 2h h 7. ■n-!B*-3|a;2+V3. Ans. inafi-T^x. 8. Differentiate (a) Vot — 16^2 with, respect to t ; (6) a + bs+ cs^ with respect to s ; (c) .Qlly^ — 8.15m2/^ — .'dim with respect to y. 9. Find the slope of the curve 42/ = a;^ — Bk — 1 at the point (1, — 2). Ans. — 1. 10. At what angle does the curve ^y = AiX — a? cut the negative axis of a; ? 32 CALCULUS 11. At wtat angles do the curves y = 3? and y = oi? intersect ? Ans. 0° and 8° 7'- 12. At what angles do the curves y = a? — Zx and y = x intersect? Ans. ges-BC and 38° 40'. 7. General Formulas of Differentiation, Continued. Theorem III. The derivative of a product is given by the formula : (III) D^uv) = uD^v + vD^u. Let 2/ =^ Mv. r Then 2/o = Mo^) 2/0 + % = (mo + ^«) (''o + ^i>), Ay = UyAv + VqAu + AmAv, Av Av , Am , . Au -^ = Wo 1- ""o 1- Am — . Ax Aa; Aa; Ax and, by Theorem I, § 5 : lim— ^= lim( ?to — |+ Hmf -Wo — )+ lim( Am — )• ^lioAa; Ax^\ Ax J i^3:^\ Ax J Axiio\ Ax J By Theorem II, § 5, the last limit has the value 0, since lim Am = and lim (Av/Asc) = Dj). The first two limits have the values UgD^v and VgD^u respectively.* Hence, dropping the subscripts, we have : D^y = uD^v + vDji, q. e. d. By a repeated application of this theorem the product of any number of functions can be differentiated. When more * More strictly, the notation sliould read liere, before the subscripts are dropped : [DiO]js=5«||, etc. Similarly in the proofs of Theorem I, II, andV. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 33 than two' factors are present, the formula is conveniently writ- ten in the form : (1) D^(uvw) _Dji ^ D^v ^ Djjo uvw u V w For a reason that will appear later, this is called the loga- rithmic derivative of uvw. Theokem IV. The derivative of a quotient is given by the formula :* (IV) -0 _ vD^u — uDj) y2 Let u y = — V Then ^^0=^ y, + ^y = '^ Vq Vq + Av . _ Mo + Am _ Mo _ 'WqAm — u^^v Vo + AlJ t'o Va(ya + £^v) ' Am Aw Aj/ Az Ax Ax Vq{vo + Av) By Theorem III of § 5 we have : limfvo^-uo^) /^x^Ax iim[vo(wo + ^«)] " Applying Theorems I and II of § 5 and dropping the sub- scripts we obtain : n „, vDm — uD.v , D^y = — ^ — I ^, q. e. d. r * The student may find it convenient to remember this formula by putting it into words: "The denominator into the derivative of the numerator, minus the numerator into the derivative of the denominator, over the square of the denominator." 34 CALCULUS Example 1. Let _ 2 — 3a; ^~l-2a!' Then i}y- a-2x)DJ2-3x)-(2-3x)DAl-2x) '^ (1 - 2xy ^ (l-2a!)(-3)-(2-3a!)(-2) ^ 1 (l-2a;)2 (l-2xy' Example 2. To prove that the theorem X>j.a;" = ma;""' is true when n. is a negative integer, n = — m. Here „ 1 a!" = — a;" TT n „ a!"i)^l — liJjB" wia!"""i _„_1 Hence Djc" = ^-— = = — nuc "" \ On replacing m in this last expression by its value, — n, the proof is complete. EXERCISES Differentiate the following functions : -^.^ -v '^ " 1 + x^ '^ 0- + x^y ■4- ^y^ ' _- v+ ' „ a^ ^ r. 3«2 — 2«' 1 — a; (1 — a!)2 5. s = Ans. Dfi = ■ ~ 1 + t ' ' (i + ty t±^. Ans ^^ + 208-0" » + a ' f? + 2az-ira?' DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 35 2 ay ax+b 9. ■:lslji.. 10. a2-2/2 a^ + a^ x + a a + bx + ciB' X 3 — 405 + a;3 ' a? +px + q j2 "-^-"-r-^ . 13 ^±1. a!2 a? 8. General rormulas of Differentiation, Concluded. Oomr posite Functions. Theobbm V. If u is expressed as a function of y and y m turn as a Junction ofx: u=fQ/), y = 4,{x), then (V) D^u = D^u-D,y. Here ya = 4> ipo), "o =/(2/o) , yo + l^y = <j>(xo + A*), Ug + ^u =f(2lo + ^V), A«=/(2/o + A2/)-/0/o), ^M - /(yo + %) -/(yo) . :^. Ax Aj/ Aa; When Aa; approaches 0, A?/ also approaches 0, and hence the limit of the right-hand side is Ai^ /(yo + Ay)-/(yo)Y ^^ ^] = DJ{y)D^y. \/!lv^ Ay J\^^ AxJ The limit of the left-hand side is Dji. Consequently Dji = D^u-Dj/, q.e.d. This equation can also be written in the form : (V) Dj^ = DJ{y)DM«'). 36 CALCULUS The truth of the theorem does not depend on the particular letters by •vrhich the variables are denoted. We may replace, for example, xhj t and yhj x: DfU = D^u D,x. Dividing through by the second factor on the right, we thus obtain the formula : (V") A« = §^- D,x Example 1. In § 4 we differentiated the function Va;, and we saw that other radicals can be differentiated in a similar manner. But each new differentiation required the evaluation of lim Ay/Aa; by working through the details of a limiting process. Theorem V enables us to avoid such computations, as the following example will show. To differentiate the function u = Va^ — «'. Let y = a^ — x*. Then u = Vy, and the differentiation thus comes directly under Theorem V, if we set _ ■/(y) — ^y> ^ (»') = a^ — a;^- Hence we have : (1) D,u = D^y/y D^a^ - x^). Now, the formula jy -^~_ 1 does not mean that the independent variable must be denoted by the letter x. If the independent variable is y, the formula reads : _ ^ D,-s/~y = -^- 2V^ DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 37 Consequently (1) can be written in the form : (2) D^ri = -!-(- 2 x) = - "^ 2V2/ V02- We have, then, as the final result : D,-\/a? — x^ = - Va^ — x'^ Exam/pie 2. To differentiate the function ^ (1 - x/ Let z=:l — X. Then y = z~K To apply Theorem V in the present case, the letters u and y must be replaced respectively by y and z. Thus Theorem V reads here: D^y = D^Dj^, or Dj) = D^^D^a-x). Since Formula (8) of § 2 has been extended to negative in- tegral values of n by § 7, Ex. 2, we have : Z)^==-3«-*. Hence D,y = -Zz-\-l)=-. z or D, = (l-a!)8 (1-a;)* •x-' EXERCISES Diiferentia|e A^^lowing functions : .-• V 1. (f'i ' v=Va2 + a;''. ^ras, a; P> Va2 + a;2 2.* 2/ = — ■ .dns. a; Va2 — a;2 V(a2 — a;^)^ * Note that Formula (8) of § 2 has also been shown to hold for the case n =— ^ ; § 4, Ex. 1. 38 / CALCULUS 3. 2/=Vl + a! + ai2. Ans. — "^ "^ 2 VI + iB 4- «!« A 1 < A 1 — 2a; 4. y = , . Ans. V3-2a; + 4a!2 VS — 2 a; + 4 a;^ (1 - a;)' (1 - x)* ~ x^ + 1 .6 + 4a! 6. u — — • Ans. {2-3a!)2 (2-39!)' ^ (l+2a!)* ^ (2 + a;)3 9. y=(-^Y. 10. M = ; "^ -(r^T 1 — 2a; + ai2 11.* M = a;(l — a;)<. Ans. (1 — 5a!)(l — a!)3. 12. M = a;(a + te)". -4ns. [a + (w + l)6a;](a + 6a;)"~'- 13. u = x\a + 6a;)". 14. u = a;'(l — x)*. 15. M = a;Va — a;. ^ns. ~ ^ ■ 2 Va — X 16. u = 7?^(^ — ^. 17. M = a;Vl+a; + a!2. 18. M= "^ - 19. « = - "^ Va" - a;2 Vl + a; + a;^ 20. «=JM^. 21.t ^*=^ i ^^c + d* (a2-2aa;)< 22. M = r — -• 23. M = - (a;2 — 1)2 1 + a; + a;' a;*— 36a;2+362a!-6» „_ a + 6 24. M = ^ ■ • 25. M = ■2 6 — a; {a-irhxY * Use Theorem HI. i t Do not use Theorem IV. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 39 9. Differentiation of Implicit Algebraic Functions. Wlien X and y are connected by such a relation as a;2 + y2 = a% or a^ — 2xy + 2/^ = 0, or xysaiy = x + y log x, i.e. if y is given as a function of x by an equation, F{x, y)=0 or $ (a;, y) = * (a;, y), which must first be solved for y, then y is said to be an implicit function of x. If we solve the equation for y, thus obtaining the equation y =/(»), 2/ thereby becomes an explicit function of x. By an algebraic function of a; is meant a function y which satisfies an equation of the form O{x,y)=0, ■where G(x, y) is an irreducible polynomial in x and y; i.e. a polynomial that cannot be factored and written as the product of two polynomials. Thus the polynomials are algebraic functions ; for if y = ao + aiX+ ■•• + «„»" = P{vi), then y satisfies the algebraic equation G{x,y)=y-P{x)=0. Similarly, the fractions in x are algebraic functions ; for if P(xy w^here -P(a;) and Q(<c) are polynomials having no common factor, then y satisfies the algebraic equation G{x,y) = Qix)y-Fix)=.0. 40 CALCULUS The polynomials and the fractions are also called rationed functions. Thus, ^ ■' ax + by a;2 + y^ is a rational function of the two independent variables x and y. Agaia, all roots of polynomials, as y = Vl + X + sfi, or such functions as ^1 — X are algebraic, as is seen on freeing the equation from radicals and transposing. The converse, however, — namely, that every algebraic function can be expressed by means of rational func- tions and radicals, — is not true. In order to differentiate an algebraic function, it is sufB.cient to differentiate the equation as it stands. Thus if (1) x^ + y^ = a?, we have (2) Djc''- + Dy = D,a?. To find the value of the second term, apply Theorem V, § 8. Thus Z>^2 = D^'D^ = 2yDj/. This last factor, D^, is precisely the derivative we wish to find, and it is given by completiug the differentiations indi- cated in (2) : o , o t^ n and solving this equation for Djy : y The final result is, of course, the same as if we had solved equation (1) for y : , ^ y= ± Vo* - x\ DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 41 and then differentiated : Va2 — a? y In the case, however, of the equation (3) a?-2xy + y^ = 0, we cannot solve for y and obtain an explicit function expressed in terms of radicals. Nevertheless, the equation defines y as a perfectly definite function of x ; for, on giving to x any special numerical value, as x = 2, we have an algebraic func- tion for y, — here, ^ . „ „ and the roots of this equation can be computed to any degree of precision. To find the derivative of this function, differentiate equation (3) as it stands with respect to x : (4) Dj>?-2DXxy) + D^y^ = 0. The second term in this last equation can be evaluated by Theorem III of § 7 : „ , , I>,(xy) = xD^y + y, where Dj^ denotes the derivative we wish to find. To the evaluation of the third term in (4) Theorem V of § 8 applies: Dy = 5y^Dj,. ^^^'^^ 3a!2 - 2 xDj^ -2y + 5y*D,y = 0. Solving this equation for Dj/, we have as the final result : '^ 5y*-2x Thus, for example, the curve is seen to go through the point (1, 1), and its slope there is (-D^)a,i)=-i- 42 CALCULUS The differentiation of implicit functions as set forth in the above examples is based on the assumptions a) that the given equation defines y as a function of a; ; 6) that this function has a derivative. The proof of these assumptions belongs to a more advanced stage of analysis. In the case, however, of the equations we meet in practice, — for example, such equations as come from a problem ia geometry or physics, — the condi- tions for the existence of a solution and of its derivative are fulfilled, and we shall take it for granted henceforth that this is true of the implicit functions we meet. Derivative of x", n Fractional. We are now in a position to prove the theorem „ , Djc" = nx"''- for the case that m is a fra,ction. Let q where p, q are whole numbers which are prime to each other. Let , y = »'.., Then y = x". Differentiating each side of this equatiqn with respect to x, we have : ^ ^ D,r = Djx", and since, by Theorem V, § 8, D^ = Dj>Djj = qy-^Dj), it follows that , _ qy-Wjj=px!^\ or X>^ = £-— • qy 1 This last denominator has the value (a?Y-^ = x '. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 43 Hence We see, then, that Ds = ■.naf' •1 ? q. e. d. If, finally, n is a, negative fraction, n = — m, the proof can be given precisely as was done in § 7, Ex. 2. Thus the theorem is now established for all commensurable values of n. The theorem is true even when n is irrational, e.g. m = jr or V2; the proof depends on the logarithmic function and will be given when that function has been differentiated. Example. Differentiate the function y = Va^ — a?. ' Apply Theorem V, § 8, setting z = a^ — x^. Then Z>^2/ = !>.»*= D^^D^a = |a"*(- Sk^). Hence .D^-\/a? — a^ : V(a' - 3?f EXERCISES 1. If 2!B'-3a!2t/ + 4a!2/ + 62/' = 0, findD.2/. Ans. j y ^ 6a;^ - 6a^ + 4y ^ '^ 3a;2 + 4a! + 18yJ 2. If 2/* — 2 xy'^ = a^, find D^. 3. Show that the curve x'^ — 2xy'^+'f + Zx — Sy = cuts the axis of x at the origin at an angle of 46°. 44 CALCULUS 4. Plot the curve a^ + ^ = 81, taking 1 cm. as the unit. Show that this curve is cut orthog- onally by the bisectors of the angles made by the coordinate axes. Differentiate the following functions : 5. M = Vl — X. Ans. 1 6. u = 'Va'' — 2ax + x\ Ans. 7. M = Vc*— 3c2a!-|-3ca;2- a^. Ans. — ■ 5 V(l - xy -2 3-s/a-x 3 5-Vc^ — 2cx + x^ u = \ "^ ■ 9. u = x-i/a + bx + (x>?. M — a; 10. M = ..<l^^±^. Ans. 3 Va!2(l - xy J J Va-gj+Va + a; ^^^ g^ + aVa^-a;" ^ ' a!2 Va^ — ofi 13. r=Va6. lO^a;!' 16. M = a; V2 a;. V a — a; — V a + a; 12. yrrz-v'aS^. 14. l-a,-i u= ■ x^ 15. y= Z- ■ ■VX 17. ■Vt 18. (i/^ + l)^f-y. 19. («" - o2)^ Ans. -2 2 . 2-Vf^ . 3aVs2 — o2 8* DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 45 20. "-'^ ■ 21. -''' ^-=^ V2aa!-a!2 ^ (1 + «)' 22. « = iB(a2 — x^y. 23. M = (6 — «) V& + «. 24. !Pind the slope of the curve y = x^ in the point whose abscissa is 2. Ans. tan t = .116. 25. If pv^* = c, find D^p. 26. If2/V5 = l+a;, find D^j/- ^ns. ^-=4- 2a;Va! .27. Differentiate y in two ways, where xy + A:y = Zx, and show that the results agree. - ' 28. The same, when 2/^ = 2 mx. 29. Show that the curves 32/ = 2a; + a;y, 2?/ + Sx + j/^ = aj'j/, intersect at right angles at the origin. 30. Find the angle at which the curves 2x = x'^—xy + a^, x' + y* + 5x = 7y, intersect at the origin. Ans. tan <^ = 1.4. CHAPTER III APPLICATIONS 1. Tangents and Normals. By the tangent line, or simply the tangent, to a curve at any one of its points, P, is meant the straight line through P X whose slope is the same as that of the curve at that point. Let the coordinates of F be denoted by (xq, y^. Now, the equation of the straight Fro. 9 line through P, whose slope is \, is y — ya = K^ — ai))- On the other hand, the slope of the curve at any point is Djf. If we denote the value of this slope at (a!o, ^o) ^7 {P^\, this will be the desired value of A. : Hence the equation of the tangent to the curve y=f(x) or F{x,y) = at the point (aJo, y^ is (1) y-yo= (Ps)aip - Xo). Since the normal is perpendicular to the tangent, its slope, X', is the negative reciprocal of the slope of that line, or 1 X'=- 46 APPLICATIONS 47 Hence the equation of the normal to the curve at (x^, y^) is (2) 2^-2^o=-Tp-y(aJ-!eo) or x-Xo+iDjy)o- {y -yo)=0. Example 1. To find the equation of the tangent to the curve y-a? in the point x = ^,y — \- Here Hence the equation of the tangent is y — i = ^(x — ^) or 3a! — 4?/ — 1 = 0. Example 2. Let the curve be an ellipse : t + t = l a^ V Differentiating the equation as it stands, we get ; a^ b^ ah) Hence the equation of the tangent is This can be transformed as follows : a^o2/ - aW = - 6%» + ^'^^, V%fB + ahf^y = oV + 6V = a^l^, 48 CALCULUS EXERCISES J 1. Find the equation of the tangent of the curve y = a? — X at the origin ; at the poiat where it crosses the positive axis of X. Ans. x + y = 0; 2x — y — 2 = Q. 2. Find the equation of the tangent and the normal of the circle a;2 + y2 = 4 at the point (1, VS) and check your answer. 3. Show that the equation of the tangent to the hyperbola at the point {xq, yg) is a" ¥ 4. Show that the equation of the tangent of the parabola y'^ = 2mx at the point (xo, yg) is VoV = m(» + ai))- 5. Show that the equation of the tangent of the parabola y^ = m? — 2mx at the poiat (xg, yg) is ygy = m? — m{x +Xg). 6. Show that the equation of the tangent of the equilateral hyperbola xy =ia^ at the point (xg, yg) is y^ + x^ = 2a''. 7. Find the equation of the tangent to the curve a? + y^ = a%x — y) at the origin. Ans. x = y. APPLICATIONS 49 8. Show that the area of the triangle formed by the coordi- nate axes and the tangent of the hyperbola at any point is constant. 9. Find the equation of the tangent and the normal of the curve . . „ a? = a'y' in the point distinct from the origin in which it is cut by the bisector of the positive coordinate axes. 10. Show that the portion of the tangent of the curve xi + yi = a^ at any point, intercepted between the coordinate axes, is constant. 11. The parabola y^ = 2 ax cuts the curve a^ — 3 axy + ?/'== at the origin and at one other point. Write down the equa- tion of the tangent of each curve in the latter point. 12. Show that the curves of the preceding question intersect in the second point at an angle of 32° 12'. 2. Maxima and Minima. Problem. From a piece of tin- 3 ft. square a box is to be made by cutting out equal squares from the four corners and bending up the sides. Detei;-. mine the dimensions of the box of this description which will hold the most. ^ ^' }- 3— 2a; 3— 2a; Fio. 10 Solution. Let x be the length of the ■ side of the square removed; then the dimensions of the box are as indicated in the diagrams, noting the cubical content of the box by u, we have : De- 50 CALCULUS 1) u = x(3-.2xy, 2) u = 9x — 12a;'' + 4*'. The problem is, then, to find the value of x which makes u as large as possible, x being restricted from the nature of the case to being positive and less than f : 3) 0<a!<f. , The problem can be treated graphically by plotting the curve 1). We wish to find the highest point on this curve. It appears to be the point for which x = ^, u = 2, since other values of x which have been tried lead to smaller values of u. The foregoing method has the advantage that it is direct, for it assumes no knowledge of mathematics beyond curve plotting. It has the disadvantage that curve plotting, even in the simplest cases, is labori- ous ; and, furthermore, we have not really proved that x = ^ is the best value. We have merely failed to find a better one. The Calculus supplies a means of meeting both the difficul- ties mentioned, and yielding a solution with the greatest ease. The problem is to find the highest poiat on the curve. At this point, the tangent of the curve is evidently parallel to the axis of x. Consequently, the slope of the tangent, i.e. tan T = D^u, must have the value here : Fig. 11 APPLICATIONS ^ 51 All we need do, therefore, is to compute D^u, most con- veniently from equation 2), and set the result equal to : Bji = 9 - 24a! + 12a;2 = 0. On solving this quadratic equation for x, we find two roots, •'' . — "J* T- Only one of these, however, lies within the range 3) of possible values for x, namely, the value x = ^, and hence this is the required value. EXERCISES 1. Work the foregoing problem for' the case that the tin is a rectangle 1 by 2 ft. Plot accurately the graph, taking 10 cm. as the unit, and determine in this way what appears to be the best value for x, correct to one eighth of an inch. Solve the problem by the Calculus, and show that the best value for x is .21132 ft., or 2.5359 in. 2. A farmer wishes to fence off a rectangular pasture along a straight river, one side of the pasture being formed by the river and requiring no fence. He has barbed wire enough to build a fence 1000 ft. long. What is the area of the largest pasture of the above description which he can fence off ? 3. Show that, of all rectangles having a given perimeter, the square has the largest area. 4. Show that, of all rectangles having a given area, the square has the least perimeter. 5. Each side of a shelter tent is a rectangle 6 X 8 ft. How must 1;he tent be pitched so as to afford the largest amount of room inside ? The ~^^ ^ ends are to be open. Ans. The angle along the ridge-pole must be a right angle. 52 CALCULUS 6. Divide the number 12 into two parts such that the sum of their squares may be as small as possible. (What is meant is such a division as this : one part might be 4, and then the other would be 8. The sum of the squares would here be 16 + 64 = 80.) 7. Divide the number 8 into two such parts that the sum of the cube of one part and twice the cube of the other may-be as small as possible. 8. Divide the number 9 into two such parts that the product of one part by the square of the other may be as largd^^ as possible. 9. Divide the number 8 into two such parts that the product of one part by the cube of the other may be as large as possible. 10. At noon, one ship, which is steaming east at the rate of 2^ mUes an hour, is due south of a second ship steaming south at 16 miles an hour, the distance between them being 82 miles. If both ships hold their courses, show that they will be nearest to each other at 2 p.m. 11. If, in the preceding problem, the second ship lies to from noon till one o'clock, and then proceeds on her southerly course at 16 miles an hour, when will the ships be nearest to each other? 12. Find the least value of the function y = x^ + 6x + 10. Ans. 1. cyi3. What is the greatest value of the function y = 3x — a^ for positive values of a; ? -14. -For what value of x does the function 12Vg l + 4a; attain its greatest value ? Ans. x = ^ APPLICATIONS 53 15. At -what point of the interval a < a; < 6, a being posi- tive, does the function attain its least value ? {x— a){h — x) Ans. X = Va6. 1 C*^ 16. Find the most advantageous length for a lever, by means of ■which to raise a weight of 490 lb. (see Fig. 13), q if the distance of the weight from the fulcrum is 1 ft. and the t \qx lever weighs 6 lb. to the foot. „ ° Fig. 13 i^^ 3. Continuation: Auxiliary Variables. It frequently, — in fact, usually, — happens that it is more convenient to formu- late a problem if more variables are introduced at the outset than are ultimately needed. The following examples will serve to illustrate the method. Example 1. Let it be required to find the rectangle of greatest area which can be inscribed in a given circle. It is evident that the area of the rectangle will be small when its alti- tude is small and also when its base is short. Hence the area will be largest for some intermediate shape. Let u denote the area of the rec- tangle. Then (1) M = 4 xy. But X and y cannot both be chosen arbitrarily, for then the rectangle will not in general be inseriptible in the given circle. In fact, it is clear from the Pythagorean Theorem that x and y must satisfy the relation : (2) x^ + y^ = A We could now eliminate y between equations (1) and (2), thus obtaining u in terms of x alone ; and it is, indeed, im- Fio. 14 54 CALCULUS portant to think of this elimination as performed, for there is only one independent variable ia the problem. The graph of u, regarded as a function of x, starts at the origia, rises as x increases, but finally comes back to the axis of x again when x — a. All this we read off, either from the meaning of u and x in the problem or from _* equations (1) and (2). J. „ It is better, however, in practice not to elimi- nate y, but to differentiate equa;tions (1) and (2) with respefct ta x as they -Itand, and then set D,,u = 0. Thus from (1), " ") D^u = 4:(^+xD,y)=0, and from (2), 2a; + 2 yD,y = 0. From the second of these equations we see that y Substituting for Dj) this value in the first equation, we get : y =0 or y^ = a?. y Since x and y are both positive numbers, it follows that y = x. Hence the maximum rectangle is a square. <X EXERCISES 1. Work the same problem for an ellipse, instead of a circle. 2. Work the problem for the case of a variable rectangle inscribed in a fixed equilateral triangle. Example 2. To find the most economical dimensions for a tin dipper, to hold a pint. APPLICATIONS 55 Here, the amount of tin required is to be as small as pos- sible, the content of the dipper being given. Let u denote the surface, measured in square inches. Then a) M = 2irrh + irr^. But r and h caimot both be chosen arbitrarily, for then the dipper would not in general hold a pint. If V denotes the given volume, measured in cubic Fig. 16 inches, then, since this volume can also be expressed as jrr^, we have b) irr^A = V. Differentiating equation o) with respect to r and setting D,ii = 0, we have : D,u = ir\2h + 2rDJi + 2 r j = 0, or c) h + rD,h + r = 0. DifEerentiating 6) we get : d) T\2rh + rWM=0. Now, r cannot = in this problem, and so we may divide this last equation through by r, as well as by ^ : e) 2h + rD,h, = 0. It remains to eliminate D^h between equations c) and e). From e), r Substituting this value of D,A in c), we find : /) h—2h + r = 0, or r = h. Hence the depth of the dipper must just equal its radius. Discussion. Just what have we done here ? The steps we have taken are suggested clearly enough by the solution of 56 CALCULUS Example 1. We have chosen one of the two variables, r and h, as the independent variable (here, r); differentiated the function u, which is to be made a minimum, with respect to r, and set D,m = 0. Then we differentiated the second equation 6), likewise with respect to r, eliminated D,u, and solved. But what does it all mean ? What is behind it all.? Just this : the quantity m, m the nature of the case, is a function of r. For, when to r is given any positive value, a dipper can be constructed which will fulfill the requirements. Now, if r is very large, we shall have a shallow pan, and evi- dently the amount of tin required to make it will be large ; — i.e. u will also have a large value. But what if r is small ? We shall then have a high cylinder of minute cross sections, i.e. a pipe. Is it clear that m, the surface, will be large in this case, too? I fear not, for it' is purely a relative question as to how high such a pipe must be to hold a pint, and I see no way of guessing intelligently. By means of equation 6), however, we see that and if we substitute this value ma), we get Ml / „ F ■ 2F Ttr^ r From this last formula it is clear that, when r is small, u actually is large. The graph of u, regarded as a func- tion of r, is therefore ia character as !^ shown by the accompanying figure. It is a continuous curve lying above the axis of r, very high when r is small, and also very high when r is large. It has, therefore, a lowest point, and for this value of r, the area u of the dipper will be least. But at this lowest point the slope of the curve, i),M, has the value 0. Thus we see, first, that we have a genu- O APPLICATIONS 57 ine minimum problem ; — there is actually a dipper of small- est area. Secondly, equations c) and d) must hold, and since from these equations it follows by elimination that r = ft, there is only one such dipper, and its radius is equal to its altitude. The problem is, then, completely solved. We inquired merely for the shape of the dipper. If the size had been asked for, too, it could be found by solving b) and /) for r and h, and expressiag Via cubic inches : F= ^^ = 28.875, fl-r-s = 28.87, ' r = 2.095. It can happen in practice that a function attaias its greatest or its least value at the end of the interval. In that case, the derivative does not have to vanish. Usu- ally, the facts are patent, and so no special investigation is needed. But it is neces- sary to assure oneself that a given problem which looks like one of the above does not come under this head, and this is done, as in the cases discussed in the text, by show- ing that near the ends of the interval the values of the func- tion are larger, for a minimum problem, than for values well withia the interval. EXERCISE Discuss in a similar manner the best shape for a tomato can which is to hold a quart. Here, the tin for the top must also be figured in. Show that the height of such a can should be equal to the diameter of the base. As to the size of the can, its height should be 4.19 in. A General Remark. It might seem as if the method used in the solution of the above problems were likely to be insecure, since the graph of such a function u might, in the very next problem, look like the accompanying figure. In such a case, there would be several values of x, for each of which D^u = 0, 58 CALCULUS and we should not know which one to take. Curiously enough, this case does not arise in practice, — at least, I have never come across a physical problem which led to this difficulty. In problems like the above, there must be at least one X for which Djt = ; and when we solve a given problem, we actually find only one x which fulfills the con- Fio. 19 dition. Thus there is no ambiguity. EXERCISES 1. A 300-gallon tank is to be built with a square base and vertical sides, and is to be lined with copper. Find the most economical proportions. Ans. The length and breadth must each be double the height. T~ 2. Find the cylinder of greatest volume which can be inscribed in a given cone of revolution. ji Ans. Its altitude is one-third that of the cone. I 3. What is the cylinder of greatest convex I_ surface that can be inscribed in the same cone ? ^" Ans. Its altitude is half that of the cone. Fio. 20 4. Of all the cones which can be inscribed in a given sphere, find the one whose lateral area is greatest. Ans. Its altitude exceeds the radius of the sphere by 33^ % of that radius. 5. Find the volume of the greatest cone of revolution which can be inscribed in a given sphere. 6. If the top and bottom of the tomato can considered in the Exercise of the text are cut from sheets of tin so that a regular hexagon is used up each time, the waste being a total loss, what wiD then be the most economical proportions for the can ? APPLICATIONS 59 7. If the strength, of a beam is proportional to its breadth and to the square of its depth, find the shape of the strongest beam that can be cut from a circular log. Ans. The ratio of depth to breadth is V2. 8. Assuming that the stiffness of a beam is proportional to its breadth and to the cube of its depth, find the dimensions of the stiffiest beam that can be sawed from a log one foot in diameter. 9. What is the shortest distance from the poiut (10, 0) to ' the parabola w2 = 4a;9 a / ■ 10. What points of the curve are nearest (4, 0) ? 11. A trough is to be made of a long rectangular-shaped piece of copper by bending up the edges so as to give a rec- tangular cross-section. How deep should it be made, in order that its carrying capacity may be as great as possible ? 12. Assuming the density of water to be given from 0° to 30° C. by the formula p = po(l + ca + pt^ + yf), where po denotes the density at freezing, t the temperature, and a=5.30xl0-^ /8 = - 6.53 X 10-6, y= 1.4x10-8, show that the maximum density occurs at t = 4.08°. 13. Tangents are drawn to the arc of the ellipse E! + ^ = 1 which lies in the first quadrant. Which one of them cuts off from that quadrant the triangle of smallest area ? 14. Work the same problem for the parabola y^ = a^ — iax. 60 CALCULUS 15. Show that, of all circular sectors having the same perim- eter, that one has the largest area for which the sum of the two straight sides is equal to the curved side. A y o o "Fia. 21 4. Increasing and Decreasing Functions. The Calculus affords a simple means of determining whether a function is in- creasrag or decreasing as the independent variable increases. Since the slope of the graph is ^iven by D^, we see that when D,y is positive, y iacreases as X iacreases, but when Dj/ is negative, y decreases as x ia- creases. Figure 21 shows the graph ia general when D^ is positive. In each figure both x and y have been taken as positive. But what is said above in the text is equally true when one or both of these variables are negative; for the words increase and decrease as here used mean algebraic, not numerical, ia- crease or decrease. Thus if the temperature is ten degrees below zero (i.e. — 10°) and it changes to eight below (— 8°), we say the temperature has risen. If we measure the time t, in hours from noon, then 10 a.m. will correspond to t = — 2. Let u denote the temperature, meas- ured iu degrees. Then a tem- perature chart for 24 hours from midnight to midnight might look like the accompanying figure. At any instant, t = t', for which the slope of the curve, D,u, is positive, the temperature is rising, no matter whether the thermometer ia above zero or below, and no matter whether t is positive or negative ; and similarly, when Z>,m is negative, the temperature is falling. Again, suppose the amount of business a department store t = lZ ^o. 22 APPLICATIONS 61 does in a year, as represented by the net receipts each, day, be plotted as a curve (y = receipts, measured in dollars ; x = time, measured lq days), the curve being smoothed ia the usual way. Then a point of the curve at -which the derivative is positive (i.e. D^y > 0) indicates that, at that time, the busi- ness of the firm was increasing ; whereas a point at which D^ < means that the business was falling off. We can state the result in the form of a general theorem, .the proof of which is given by inspection of the figure (Fig. 21) and the other forms of the figure, brought out in the above discussion. Theorem : When x increases, then (a) if Djf > 0, y increases; (b) if ' ' D^y < 0, y decreases. Application. As an application consider the condition that a curve y =f{x) have its concave side turned upward, as ia Fig. 23. The slope of the curve is a function of x : FlQ. 23 tan T = ^(w). For, when x is given, a point of the curve, and hence also the *-* slope of the curve at this point, is determined. Consider the tangent line at a variable point P. If we think of P as tracing out the curve and carrying the tangent along with it, the tangent will turn in the counter clock-wise sense, the slope thus increasing algebraically as x increases, whenever the curve is concave upward. And con- versely, if the slope increases as x increases, the tangent will turn in the counter clock-wise sense and the curve will be con- cave upward. Now by the above theorem, when Z>^tanT>0, tauT increases as x increases. Hence the curve is concave upward, when D, tan t is positive ; and conversely. 62 CALCULUS The derivative D^tanr is the derivative of the derivative of y. This is called the second derivative of y, and is denoted as follows : ^ . „ . -r, „ (read : " D x second of y ").* The test for the curve's being concave downward is obtained in a similar manner, and thus we are led to the following important theorem. Test fob a Cukve's being Concave Upward, etc. The curve ,, , is CONCAVE UPWARD whsn D^ > ; CONCAVE DOWNWARD when D^ < 0. A point at which the curve changes from being concave upward and be- comes concave downward (or vice versa) is called a poirU of inflection. Since D^y changes sign at such a point, this function will' necessarily, if continuous, vanish there. Hence : A necessary condition for a point of inflection is that BJ'y = 0. Example. Consider the curve y = 3? — 3x. * The derivative of the second derivative, Dx^DJiy), is called the third derivative and is written D^^, and so on. Fig. 24 APPLICATIONS 63 Its slope at any point is given .by the equation The second derivative of y with respect to x has the value D/2/ = 6a;. Thus we see that this curve is concave upward for all positive values of x, and concave down- ^la. 25 ward for all negative values. In character it is as shown in the accompanying figure. EXERCISES For what values of x are the following functions increasing ? For what values decreasing ? 1. 2/ = 4 — 2a;2. 2. y = x^ — 2x + Z. Ans. Increasing, when a; > 1 ; decreasing, when x < 1. 3. 2/ = 5 + 12x — a;2. 4. y=3fi — 21x + l. Ans. Increasihg, when as > 3, and when a; < — 3 ; decreas- ing, when — 3 < a; < 3. ^. y = 5 -\-&x — 3?. 6. y = x — x". 7. y = ofi — ^x^ + 12x — l. In what intervals are the following curves concave upward ; in what, downward ? 8. y = a^ — 3a;2 + 7a; — 5. Ans. Concave upward, when a; > 1 ; concave downward, when a; < 1. 64 CALCULUS 9. 2/ = 15 + 8a; + 3a!= — 05*. 10. y = a^ — Qx^ — x — 1. 11. y = 3 — 9x + 24:x^—ia^. 12. y = 2a>> — X*. 13. y = as* — 4.0^ — 6x + 11. 14. y ==-121x + 7x^ — x\ 15. y = 13 + 23x-24:X'' + 12a^-x^. 5. Curve Tracing. In the early work of plotting curves from their equations the only way we had of finding out what the graph of a function looked like was by computiug a large number of its points. We are now in possession of powerful methods for determining the character of the graph with scarcely any computation. I"or, first, we can find the slope of the curve at any point; and, secondly, we can determine in what intervals the curve is concave upward, in what concave downward.* Example. Let it be required to plot the curve (1) 3y = x^- 3x^ + 1. a) Determine first its slope at any point : (2) 3n^ = 3x^-6x, Djj = v? — 2x. * There are two great applications of the graphical representation of a function. One is quantitaMvie, the other, quaUtative. By the first I mean the use of the graph as a table, for actual computation. Thus in the use of logarithms it is desirable to have a graph of the function y = logio a drawn accurately for values of x between 1 and 10 ; for by means of such a graph the student can read ofi the logarithms he is using, correct to two or three significant figures, and so obtain a check on his numerical vork. Therfe is, however, a second large and important class of problems, in which the character of a function is the important thing, a minute deter- mination of its values being in general irrelevant. A case in point is the determination of the number of roots of an alge- braic equation, e.g. ss _ ^2 - 4 x -t- 1 = 0, Here, we plot the curve y ^^.^ _xi - 4:x + 1 and inquire where it cuts the axis of x. For this purpose it is altogether adequate to know the character of the curve, and for treating this problem the methods of the present paragraph yield a powerful instrument. APPLICATIONS 65 It is always useful to know the points at which the tangent to the curve is parallel to the axis of x. These are obtaiued hy setting D^y = and solving. Thus we get from (2) the equation : a;2 — 2a; = 0. The roots of this equation are x = and x = 2 Now determine accurately the points having these abscissas, plot them, and draw the tangents there : y We do not yet know whether the curve lies above its tangent in one of these points, or be- low its tangent ; it might even cross its tangent, for the point might be a point of inflection. These questions will all be answered by aid of the second derivative. 6) Compute the second derivative : DJh/ = 2a; - 2 = 2(!B - 1). We see that it is positive when x is greater than 1 and nega- tive when X is less than 1 : Fib. 26 -D.V > when l<x; D^^y < when X<1. DJh, = when x=l. Hence the curve has a point of inflection when x=l. This is a most important point on the. curve. We will compute its 66 CALCULUS coordinates accurately, determine the slope of the curve there, and draw accurately the tangent there. yU=. = -i; -D^U=i — 1. This is the last of the important tangents which we need to draw. Since the curve is concave upward to the right of the liue X = 1, and concave downward to the left of that line, it must be in character as indicated. We see, then, that it cuts the axis of x between and 1, and again ta the right of the poiut a; = 1 ; and it cuts that axis a third time to the left of the origin. These last two points can be located more accurately by computing the function for a few simple values of x. hence the curve cuts the axis of x between a; = 2 and a; = 3. 2/L=-i=-3; hence the curve cuts the axis between a; = and a; = — 1. Incidentally we have shown that the cubic equation a? — 3x^ + 1 = has three real roots, and we have located each between two successive integers. EXERCISES Discuss in a similar manner the following curves. In par- ticular : a) Determine the points at which the tangent is horizontal, if such exist, and draw the tangent at each of these points ; 6) Determine the intervals in which the curve is concave upward, and those in which it is concave downward ; c) Determine the points' of inflection, if any exist, and draw the tangent in each of these points ; APPLICATIONS 67 d) Draw in the curve.* In most cases it is desirable to take 2 em, as the imit. 1. y = x^ + 3x^ — 2. 2. y = a^—3x + l. 3. y=iifl +3x + l. 4. 6y = 2a^-3x^-12x + 6. 5. 6y = 2d' + 3x^-12x — 4:. 6. y = a? + x^ + x + l. Suggestion. Show that the derivative has no real roots and hence, being continuous, never changes sign. 7. 12y = 4:a?-6x^ + 12x — 9. 8. y = 2x' — x — a^. 9. 12y = 4a!8 + 18a!2 + 27a; + 12. 10. 3/ = l — 4a! + 6c(;2_3ffi3. 11. y = l + 2x + x^ — 0(?. 12. 42/ = a!* — 6a!2 + 8. 13. y = a^ — ^ix"^ + 1. 14. y = x — a^. 15. y = X + x^. 16. y = a!* + a;2. 17. y=2xl*~xK 18. y = 3a!5 + 6a!' + 15a! + 2. 19. 60y = 23^ + 16a^ + GOaj^ — 30. 6. Relative Maxima and Minima. Points of Inflection. A function (1) • 2/ =/(«') * Since a curve separates very slowly from its tangent near a point of inflection, the material graph of the curve must necessarily coincide with the material graph of the tangent for some little distance. 68 CALCULUS is said to have a maximum at a point x = Xoii its value at Xq is larger than at any other point in the neighborhood of a%. But such a maximum need not represent the largest value of the function in the complete interval a ^ a; ^ 6, as is shown by- Fig. 27, and for this reason it is called a relative maxi- mum, in distinction from a maximum maximorum, X or an absolute maximum. a;=a x„ x^ x=b A. similar definition ^°- ^ holds for a minimum, the word " larger " merely being replaced by " smaller." It is obvious that a characteristic feature of a maximum is that the tangent there is parallel to the axis of x, the curve being concave downward. Similarly for a minimum, the curve here beiag concave upward. Hence the foUowiag Test fok a Maximum ob a Minimum. If (a) [Z)^]_=0, [i).^]_„<0, the function has a maximum for x = Xo; if (6) iDjy-]^=0, [i).^]_„>0, it has a minimum. The condition is sufficient, but not necessary ; cf . § 7. Example. Let y = !ic^ — 3x'' + l. Here D^ = Gaj^- 6a; = 6a;(a!2- l)(a!= + 1), and hence D^y = for a; = — 1, 0, 1. Thus the necessary condition for a maximum or a minimum, Dj/ = 0, is satisfied at each of the points a; = — 1, 0, 1. To complete the determination, if possible, compute the second derivative, j)^^^s0<^-6, and determine its sign at each of these points : Fio. 28 APPLICATIONS 69 [i)j^]j=_i = 24 > 0, .-. x = —l gives a minimum; [D,^2/],_o =— 6<0, .•.x= gives a maximum ; [PJh/],,^ = 24 > 0, .-. x= 1 gives a minimum. Points of Inflection. A point of inflection is characterized geometrically by the phenomenon that, as a point P describes the curve, the tangent at P ceases rotating in the one di- y rection and, tumiag back, be- gins to rotate in the opposite direction. Hence the slope of the curve, tan t, has either a maximum or a minimum at — a point of inflection. Conversely, if tan r has a maximum or a minimum, the curve will have a point of inflec- tion. For, suppose tan t is at a maximum when x = Xg. Then as X, starting with the value Xq, increases, tan t, i.e. the slope of the curve, decreases algebraically, and so the curve is con- cave downward to the right of a^. On the other hand, as x decreases, tan t also decreases, and so the curve is concave up- ward to the left of X(,. Now, we have just obtained a theorem ■W'hich insures us a maximum or a minimum in the case of any function which satisfies the conditions of the theorem. If, then, we choose as that function, tanr, the theorem tells us that tanr wUl surely be at a maximum or a minimum if D,tanr = 0, DJ^ tsmr^O. Hence, remembering that tan T = D^y, we obtain the following Test fok a Point of Inflection. If the curve has a point of inflection at x— Xf,. 70 CALCULUS This test, like the foregoLag for a maxionim or a minimum, is sufficient, but not necessary ; cf. § 7. Example. Let . 272/ = ffi* + 2a!S-12a;2 + 14a; — 1. Then 27D,y = 4:afi + 6x' — 2ix + U, 27 DJh/ = 12a:'' + 12 a! - 24 = 12(a! - l)(a; + 2), 2rZ),52/ = 12(2a; + l). Setting D/y = 0, we get the points a; = 1 and x = — 2. And S1I1C6 27[A'2/]^= 36 =ifc 0, 27[X>,'2^]_., 36 =jfc 0, we see that both of these points are points of inflection. The s,lope of the curve in these points is given by the equations : 27[i?^]^i=0, 27[Z)^]_3=54. Hence the curve is parallel to the axis of x at the first of these points ; at the second its slope is 2. EXERCISES Test the following curves for maxima, minima, and points of inflectioil, and * determine the slope of the curve in each point of inflection. 1. y =: 4:0^— 15x^+12 x+1.- 3. 6y = afi — Ssc'+Sa^—l. Z. y = !i^ + a^ + iifi. 4. y=(x-iy{x+2y. . ^- y=-, 2 + 3a? 6. y=(l — x^^ 7. Deduce a test for "distinguishing be- tween two such points of inflection as those indicated in Fig. 29. APPLICATIONS 71 7. Necessary and Sufficient Conditions. In order to under- stand the nature of the tests obtained in the foregoing paragraph it is essential that the student have clearly in miad the mean- ing of a necessary condition and of a sufficient condition. Let us illustrate these ideas by means of some simple examples. a) A necessary condition that a quadrilateral be a square is that its angles be right angles. But the condition is obviously not sufficient ; all rectangles also satisfy it. 6) A sufficient condition that a quadrilateral be a square is that its angles be right angles and each side be 4 in. long. But the condition is obviously not necessary | the sides might be 6 iu. long. c) A necessary and sufficient condition that a quadrilateral be a square is that its angles be right angles and its sides be mutually equal. As a further illustration consider the following. It is a well-known fact about whole numbers that if the sum of the digits of a whole number is divisible by 3, the number is divis- ible by 3 ; and conversely. Also, if the sum of the digits of a whole number is divisible by 9, the number is divisible by 9 ; and conversely. Hence we can say : i) A necessary condition that a whole number be divisible by 9 is that the sum of its digits be divisible by 3. But the condition is not sufficient. ii) A sufficient condition that a whole number be divisible by 3 is that the sum of its digits be divisible by 9. But the condition is not necessary. iii) A necessary and sufficient condition that a whole num- ber be divisible by 3 (or 9) is that the sum of its digits be divisible by 3 (or 9). t Turning now to the considerations of § 6, we see that a necessary condition for a miaimum is that 72 CALCULUS at the point in question, a; = a^. But this condition is not sufficient. When it is fulfilled, the function may have a maximum, or it may have a point of inflection with horizontal tangent. On the other hand, the condition is sufficient for a minimum. But it is not necessary. Thus the function (1) y=x^ ohviously has a minimum when x = 0. The necessary condi- tion, Dj/ = 0, is of course fulfilled : But here Z>,=2/ = 12 a;^, and [D^^']^ is not positive ; it is 0. Again, as was shown in § 4, a necessary condition for a point of inflection is that ^ „ DJhf = at that point. But this condition is not sufficient. Thus in the case of the curve (1) this condition is fulfilled at the origin. But the origin is not a point of inflection. Remark. It may seem to the student that such tests are unsatisfactory since they do not apply to all cases and thus appear to be incomplete. But their very strength lies in the fact that they do not tell the truth in too much detail. They single out the big thing in the cases which arise in practice and yield criteria which can be applied with ease to the great majority of these cases. 8. Velocity; Bates. By the average velocity with which a point moves for a given length of time t is meant the distance 8 traversed divided by the time : average velocity = -■ APPLICATIONS 73 Thus a railroad train which covers the distance between two stations 16 miles apart in half an hour has an average speed of 15/-|^= 30 mUes an hour. When, however, the point in question is moving sometimes fast and sometimes slowly, we can describe its speed approxi- mately at any given instant by considering a short interval of time immediately succeeding the instant t^ in question, and taking the average velocity for this short interval. For example, a stone dropped from rest falls according to the law : . „ „ To-find how fast it is going after the lapse of to seconds. Here (1) So = 16«o'. A little later, at the end of t' seconds from the beginning of the fall, (2) s' = 16«'2 and the average velocity for the interval of t' — to seconds is (3) ^j^^ ft. per second. t — to Let us consider this average velocity, in particular, after the lapse of 1 second : ^ . „ to = 1, So •= 16. Let the interval of time, t' — to, be -^ sec. Then s' = 16 X 1.1== = 19.36, s;^-So _ 3^ ^ 336 ft. a second. t' - to .1 Thus the average velocity for one-tenth of a second immedi- ately succeeding the end of the first second of fall is 33.6 ft. a second. Next, let the interval of time be j^ sec. Then a similar computation gives, to three significant figures : ^^^^ = 32.2 ft. a second. t' - to 74 CALCULUS And when the interval is taken as x^^nr ^^°-' *^® average velocity is 32.0 ft. a second. These numerical results iadicate that we can get at the speed of the stone at any desired instant to any desired degree of accuracy by direct computation ; we need only to reckon out the average velocity for a sufB.eiently short interval of time succeeding the iastant in question. We can proceed in a similar manner when a point moves according to any given law. Can we not, however, by the aid of the Calculus avoid the labor of the computations and at the same time make precise exactly what is meant by the velocity of the point at a given instant? If we regard the interval of time- 1' — io as an increment of the variable t and write «' — <o = Ai, then s' — So = As will represent the corresponding increment in the function, and thus we have : average velocity = — Now allow At to approach as its limit. Then the average velocity will in general approach a limit, and this limit we take as the definition of the velocity, v, at the instant ^ : lim (average velocity from t=tgto t = f) = actual velocity * at instant t = to, or v = lim — = D,s. At^ At Hence it appears that the velocity of a point is the time- derivative of the space it has traveled. In the ease of a freely falling body this velocity is v = D,s = 32 1. In the foregoing definition, s has been taken as the distance actually traversed by the moving point, P. More generally, let s denote the length of the arc of the curve on which P is moving, s being measured from an arbitrarily chosen fixed * Sometimes called the instantarieous velocity. APPLICATIONS 75 point of tliat curve. Either direction along the curve may be chosen as the positive sense for s. Thus, in the case of a freely falling body, s might be taken as the distance of the body above the ground. If h denotes the initial distance, then where s' denotes the distance actually traversed by P a' at any given instant. Hence , D,s + D/ = 0, ^ s or D,s = — D,s'. Fig. 30 Here D^ gives numerically the value of the velocity, but D^s is a negative quantity. We will, accordingly, extend the conception of velocity, defining the velocity v of the point as D,s : V = DfS. Thus the numerical value of v or D,s will always give the speed, or the value of the velocity in the earlier sense. In case s increases with the time, DfS is positive and represents the speed. If, however, s decreases with the time, D,s is nega- tive, and the velocity, v, is therefore here negative, the speed now being given by — « or — X),s. In all eases, Speed=|u| = \D,s\. Example. Let a body be projected upward with an initial velocity of 96 ft. a second. Assuming from Physics the law that s = 96f-16«2, find its velocity a) at the end of 2 sec. 6) at the end of 5 sec. Solution. By definition, the velocity at any instant is V = D,s = 96 -32t. Hence a) v\^,=^. ^V 6) «l^ = -64. 76 CALCULUS The meaning of these results is that, at each of the two instants, the speed is the same, namely, 64 ft. a second (and the height above the ground is also seen to be the same, s = 128 ft.). But when t = 2, DiS is positive ; hence s is in- creasing with the time and the body is rising. When t = 5, D,s is negative ; hence s is decreasing with the time, and the body is descending. Rates. Consider any length or distance, r, which is chang- ing with the time, and so is a function of the time. Let j-q denote the value of »• at a given Instant, t = to, and let r' be the value of r at a later instant, t — t'. Then the increase in r will be r' — ro = Ar and that in t will be t' — ta = ^t. Thus in the interval of time of At seconds succeeding the instant t = Jj, average rate of increase of r = — At Now let At approach as its limit. Then the average rate of increase will in general approach a limit, and this limit we take as the definition of the rate of increase of r at the instant t^ : lim (average rate of increase from ^ = {^ to f = t') = actual rate of increase at instant f = <o = lim — = Dir. A«=o At In other words, the rate at which r is increasing at any in- stant is defined as the time-derivative of r. If r is decreasing, D,r wUl be a negative quantity ; and con- versely, if D,r is negative, then r is decreasing. In either ease, the numerical value of D^r gives the rate of change of r ; just as, in the case of velocities, the numerical value of D,s gives the speed. More generally, instead of r, we may have any physical quantity, u, as an area or a volume or the current in an electric circuit or the number of calories in a given body. APPLICATIONS 77 In all these cases, the rate at which u is increasing ia defined as the time-derivative of u, i.e. as D,u ; and the rate of change of M is I D,u |. Example. At noon, one ship is steaming east at the rate of 18 miles an hour, and a second ship, 40 miles north of the first, is steaming south at the rate of 20 miles an hour. At what rate are they separating from each other at one o'clock ? Solution. The relation between r and t is here given by the Pythagorean Theorem : or r' = (A0-20ty+{lSty, 1 \ (1) r2 = 1600 - 1600* + 724^2. i \ Hence 18i\ (2) r = V1600 - IGOOt + 724<2. Fig. 31 We wish to find D,r. This can be done by difEerentiatiag equation (2) ; but that would be poor technique, since it is simpler to diEEerentiate equation (1) through with respect to t: (3) 2?-i),r = -1600 + 1448<, - 800 + 724« D,r = Equation (3) gives the rate at which r is increasing at any instant t ; i.e. t hours past noon, or at t o'clock. Setting now, in particular, * = 1, we obtain : A»-|.=i=- 76 = - 2.825. The meaning of this result is twofold. First, since D,r is negative when f = 1, the ships are not receding from each other, but are coming nearer together. Secondly, the rate of change of the distance between them is, at one o'clock, 2.825 miles an hour. 78 CALCULUS Let the student determine how long they will continue to approach each other, and what the shortest distance between them will he. Remark. It is important for the student to reflect on the method of solution of this problem, since it is typical. We were asked to find the rate of recession at just one instant, t=l. We began by determining the rate of recession generally, i.e. for an arbitrary instant, t=t. Having solved the general problem^ we then, as the last step -in the process, brought into play the specific value of f which alone we cared for, namely, t = l. The student will meet this method again and again, — in integration, in mechanics, in series, etc. We can formulate the foregoing remark suggestively as follows : By means of the Calculus we can often determine a particular physical quantity, like a velocity, an area, or the time it takes a body, acted on by known forces, to reach a certain position. The method consists in first determining a function, whereby the general problem is solved for the variable case ; and then, as the last step in the process, the special numerical values with which alone the proposed question is concerned, are brought into play. EXERCISES 1. The height of a stone thrown vertically upward is given bytheformula: s=m-16t^. When it has been rising for one second, find (a) its average velocity for the next ^ sec. ; (6) for the next ^^ sec. ; (c) its actual velocity at the end of the first second ; (d) how high it will rise. Ans. (a) 14.4 ft. a second ; (6) 15.84 ft. a second ; (c) 16 ft. a second ; (d) 36 ft. 2. One ship is 80 miles due south of another ship at noon, and is sailing north at the rate of 10 miles an hour. The APPLICATIONS 79 second ship sails west at the rate of 12 miles an hour. Will the ships be approaching each other or receding from ^each other at 2 o'clock ? What -will be the rate at which the dis- tance between them is changing at that time ? How long will they continue to approach each other ? 3. If two ships start abreast half a mile apart and sail due north at the rates of 9 miles an hour and 12 miles an hour, how far apart will they be at the end of half an hour ? How fast will they be receding at that time ? 4. Two ships are steaming east, one at the rate of 18 miles an hour, the other at the rate of 24 miles an hour. At noon, one is 50 miles south of the other. How fast are they sepa- rating at 7 P.M. ? 5. A ladder 20 ft. long rests against a house. A man takes hold of the lower end of the ladder and walks off with it at the uniform rate of 2 ft. a second. How fast is the upper end of the ladder coming down the wall when the man is 4 ft. from the house ? 6. A kite is 150 ft. high and there are 250 ft. of cord out. If the kite moves horizontally at the rate of 4 m. an hour directly away from the person who is flying it, how fast is the cord being paid out ? Ans. 3^ m. an hour. 7. A stone is dropped into a placid pond and sends out a series of concentric circular ripples. If the radius of the outer ripple increases steadily at the rate of 6 ft. a second, how rapidly is the area of the water disturbed increasing at the end of 2 sec. ? Ans. 452 sq. ft. a second. 8. A spherical raindrop is gathering moisture at such a rate that the radius is steadily increasing at the rate of 1 mm. a minute. How fast is the volume of the drop increasing when the diameter is 2 mm. ? 9. A man is walking over a bridge at the rate of 4 miles an hour, and a boat passes under the bridge immediately below him rowing 8 miles an hour. The bridge is 20 ft. above the 80 CALCULUS boat. How fast are the boat and the man separating 3 min- utes later ? Suggestion. The student should make a space model for this problem by means, for example, of the edge of a table, a crack La the floor, and a string ; or by two edges of the room which do not intersect, and a string. He should then make a drawing of his model such as is here indicated. 10. A locomotive running 30 miles an hour over a high bridge '"■ dislodges a stone lying, near the track. The stone begins to fall just as the locomotive passes the point where it lay. How fast are the stone and the loco- motive separating 2 sec. later ? * 11. Solve the same problem if the stone drops from a point 40 ft. from the track and at the same level, when the locomo- tive passes. 12. A lamp-post is distant 10 ft. from a street crossing and 60 ft. from the houses on the opposite side of the street. A man crosses the street, walking on the crossing at the rate of 4 miles an hour. How fast is his shadow moving along the walls of the houses when he is halfway over ? * BScHEB, Plane Analytic Geometry, p. 230. CHAPTER IV INFINITESIMALS AND DIFFERENTIALS 1. Infinitesimals. An infinitesimal is a variable which it is desirable to consider only for values numerically small and which, when the formulation of the problem in hand has pro- gressed to a certain stage, is allowed to approach as its limit. Thus in the problem of differentiation, or finding the limit (1) lim^=Z>^, Aa! and Ay are infinitesimals ; for we allow Aa; to approach as its limit, and then Ay also approaches 0.. Again, if we denote the value of the difference Ay/ Ax — Dj/ by £, so that (2) f-2>^ = e, Ao; then e is an infinitesimal. For, when Aa; approaches 0, the left-hand side of equation (2) approaches 0, and so e is a vari- able which approaches as its limit, i.e. an infinitesimal. Principal Infinitesimal, When we are dealing with a num- ber of infinitesimals, a, p, y, etc., it is usually possible to choose any one of them as the independent variable, the others then becomiag functions of it, or dependent variables. That infinitesimal which is chosen as the independent variable is called the principal infinitesimal. Thus, if the infinitesimals are a and jS, and if 81 82 CALCULUS it is natural to choose « as the principal infinitesimal. But it is perfectly possible to take j8 as the principal infinitesimal. « then becomes the dependent variable, and is expressed ia terms of p by solving equation (3) for a. : (4) « = — ^ ^' 2-3/3 Order of Infinitesimals. We are going to separate infinitesi- mals into classes, according to the relative speed with vrhich they approach 0. Suppose we let a. set the pace, taking on the values .5, .1, .01, .001, etc. Consider, for example, a'. Then «' takes on the respective values .25, .01, .0001, etc., and hence runs far ahead of a : « .5 .1 .01 .001 ■•. oC' .25 .01 .0001 .000001 - Purthermorej the closer the two get to 0, the relatively nearer 0? is to 0. Thus, when a = .5, a^ is twice as close ; but when a = .01, «' is one hundred times as close ; and so on. Again, consider the infinitesimal \a. It is always twice as close to as a is. Similarly, 10 « is always one-tenth as close as a,. From these examples we see that there is a decided difference between the relative behavior of a and Tea on the one hand, and that of a and «2 on the other. For, lea. is keeping pace relatively with a, whereas a^ runs indefinitely ahead of «, rela- tively. Consequently, we should put lea into the same class with «, whereas a? forms the starting point for a new class. To this latter class would belong such infinitesimals as ^«' or 4a' — a' ; and the former class would include, for example, 2« + 3a- and -fj-j^a — lOOOa*. Let the student make out a table like the above for each of these examples. What is the common property of all infinitesimals of the same class ? Is it not, that, for two infinitesimals, the relative speed with which they approach is nearly, or quite, a fixed number not zero ? It is this idea which lies at the bottom of INFINITESIMALS AND DIFFERENTIALS 83 the conception of the order of an infinitesimal, and it is for- mulated in a precise definition as follows : Definition. Two infinitesimals, p and y, are said to be of the same order if their ratio approaches a limit not ; lim^=ir:^'0. V Thus ;8=2« + a? and y = 3a — a' are of the same order. For, j8^ 2ct + «' ^ 2+ tt y 3 a — a? 3 — o? and hence, when « approaches 0, ,. fl ,. 2-l-a 2 . limS=lim-^ = -=^0. Similarly, 12 o? + 5a^ and Ba^ — 7 a' are infinitesimals of the same order. An infinitesimal /3 is said to be of higher order than y if limi2 = 0, 7 Thus if j3=9a* and y = 2a + 5a*, j8 is of higher order than a. For, ^^ 9a' _ 9a y~2a+5a* 2 + 5a" and hence, when a approaches 0, ,. fl T 9a y 2 + 5 «' Finally, p is said to be of lower order than y if (5) lim"=oD, (read: "j8/y becomes infinite" ; not ";8/y equals infinity."*). * The student should now turn back to Chapter 11, § 6, and read again carefully what is said there ahout infinity. In particular, he should im- 84 CALCULUS Thus if /8=Va and y=6a + a», /8 is of lower order than y. For ;3_ V^ ^ 1 y 6«+a' Va(6 + a') When a approaches 0, it is evident that the last fraction in- creases without limit, or „ lim^ = oo. y First Order, Second Oi-der, etc. An infinitesimal /i is said to be of the ^rs< order if it is of the same order as the principal infinitesimal, a ; i.e. if „ a If /8 is of the same order as a\ i.e. if lim4 = ^¥=0, then ^ is said to be of the second order. And, generally, if ff is of the same order as a", i.e. if limJ?.= ^:^0, a" then p is said to be of the n4h order. Thus if /8 = 2a or p = ^— or Q = a + (t\ 2 — a then j8 is of the first order. But if ^ = 2«2+a' or I3 = -J^ or B=a\ 3 + a then fi is of the second order. press on his mind the fact that infinity is not a limit and that in the notation used in (5) the = sign does not mean that one number is equal to another number. The formula is not an equation in the sense in which 2x = 3 or o' — 62 = (a — 6) (a + 6) is an equation. The formula means no more and no less than that the variable /S/7 increases in value without limit. INFINITESIMALS AND DIFFERENTIALS 85 If^=Va, then /3 _ . J and Iim-^=lgfcO. a* Hence /8 is of the order ^. It is easily seen that if two infinitesimals /S and y are, under the present definition, each of order n, then they also satisfy the earlier definition of being of the same order. For, let hm.-^ = K^O and lim-^ = i:^0. «■" a" Then, if we denote the differences /S/et" — K and y/a" — L respectively by e and rj, so that (6) K-K=e and ^-i = «, «" a" these variables, e and r), will be infinitesimals. For, the left- hand side of each of the equations (6) approaches 0. From equations (6) it follows that ^ = K+ £ and ^ = Z, + ». «" a" On dividing one of these equations by the other we have : y L + rj We are now ready to allow a to approach as its limit. Then lim^=lim^^ y L + Ti By Theorem III of Chapter 2, § 5 this last limit has the value lini^±J = i-™L(:K+i) = :^. L + ri lim (L + 7i) L Hence, finally j^^ ^^^:^o, q. e. d. y L 86 CALCULUS EXERCISES 1. Snow tnat ^ = 5a— ll«2 + a3 and y = 7a + a< are infinitesimals of the same order. 2. Show that /8 = 2«-3a2 and y = 2a + a* are infinitesimals of the same order, but that their difference, ^ — y, is of higher order than yS (or y). 3. Show that B = is an infinitesimal of the second ^ a' -2 order, referred to « as_principal infinitesimal. 4. Show that /8 = Va^ + 2 a* is of the first order, referred to a. ■ 5. Show that ^ = V2« + 13a' is of lower order than a. 6. Show that the order of fi in question 5 is n = ^. Determine the order of each of the following infinitesimals, referred to a as the principal infinitesimal : 7. |a + 18a3. 11. -s/i^ - a. 8. -a+V2«' + «*. 12. ^- aW + „i8. la!' *■ 13 -a ' ^^- v2a2-«'. 10. \8-7« ■ >/ «2 + 4tt* + 2 15. If and y are infinitesimals of orders n and m respec- tively, show that their product, ^y, is an infinitesimal of order M + m. 16. If j8 and y are infinitesimals of the same order, show that their sum is, in general, an infinitesimal of the same order. Are there exceptions ? Illustrate by examples. INFINITESIMALS AND DIFFERENTIALS 87 2. Continuation ; Fundamental Theorem. Principal Part of an Infinitesimal. Let j8 be an iniiiiiteaimal of order n, and let a be the principal infinitesimal. Then lim-^ = ir^O. cC Moreover, as pointed out in the last paragraph, (1) 4=^+^' a" where e is infinitesimal. Prom (1) it follows that (2) p = Za" + ea". This last equation gives a most important analysis {i.e. break- ing up) of /3 into two parts, each of which is simple for its own peculiar reason. i) Ka'^ is the simplest infinitesimal of the nth. order imagi- nable, — a monomial in the independent variable, the function y = Kx\ ii) cot'' is an infinitesimal of higher order than the nth. The first part, ^a", is called the principal part of ^S. By far the most important case in practice is that of infini- tesimals /8 of the ^rst order, n = 1. Here «" a and /8 = Ka -\- eot. Hence we see that the principal part of an infinitesimal of the first order is proportional to the principal infinitesimal. Example 1. Let yS = 2a — a^. Then p is obviously of the first order, or m = 1, and here a" a 88 CALCULUS Clearly, theu, K=2, € = — a, and the principal part of p is 2a. Example 2. Let ;8 = 2«- 7-4a Here, obviously, n = = 2, and - lim-^ = a^ 7-4a .2 ■7' Hence K=-- 7 By definition e = a" In the present case, then. £ = - 2 , 2 8a 7-4a 7 7(7-4 ^) EXERCISE Determine the principal parts of a goodly number of the infinitesimals occurring in the Exercises at the end of § 1. Equivalent Infinitesimals. Two infinitesimals, as ^ and y, shall be said to be equivalent if the limit of their ratio is unity : lim^=l. y For example, the following pairs of infinitesimals are equiv- alent : i) 2 a 4- a2 and 2 a + a^ ; ii) \a?- — a? and \a^ + a?; iii) V2 a + 5 a2 and V2a— 7«^. An infinitesimal and its principal part are always equivalent infinitesimals. For, if Ea" is the principal part of /8, then INFINITESIMALS AND DIFFERENTIALS 89 ■where ■q is of higher order than Ka". Hence -^ 1+-^, lim-^=l + limJ-. Ka" Ka"' Ka" Ka" But lim rj/Ka" — 0, and the statement is established. Two infinitesimals which have the same principal parts are equivalent, and conversely. Equivalent infinitesimals are of the same order ; but the converse is not true. The difference between two equivalent infinitesimals, /8 and y, namely, /3 — y, is of higher order than j8 or y. For y y hence lim ^-^^ = lim f *? - l'^ y -(?-) = Aim ^V 1 = 0, q.e.d. Conversely, if (3 and y are two infinitesimals whose differ- ence, /8 — y, is of higher order than /3 or y, then (3 and y are equivalent. For, since i2^l2 = ^_l, y 7 it follows that j.^/^ _ ^\ li^^sU^. The right-hand side of this equation is by hypothesis, and the left-hand side is equal to Aim^Vl. ^ yJ Hence lim"=l, q.e.d. y We come now to a theorem of prime importance in the Infinitesimal Calculus. 90 CALCULUS Fundamental Theorem. The limit of the ratio of two infini- tesimals, 7 is unchanged if the numerator infinitesimul j8 be replaaed by any equivalent infinitesimal /3' and the denominator infinitesimal y be replaced by any equivalent infinitesimal y'. In other words : lim " = lim c. y y provided li^l^-L „^ li^^^^l. The proof is immediate. It is obvious that Hence by Theorem II, Chapter II, § 5 we have But the first and third limits on the right-hand side are each equal to 1 by hypothesis. Hence lim " = lim ^, q. e. d. y y The theorem can be stated in the following equivalent form: The limit of the ratio of two infinitesimals is the same as the limit of the ratio of their principal parts. The student must not generalize from this theorem and infer that an infinitesimal can always and for all purposes be replaced by an equivalent infinitesimal. Thus if P — 2a + a^ and y = 2a — a\ their difference, /8 — y = «' + «*, lim-'" INFINITESIMALS AND DIFFERENTIALS 91 is an infinitesimal of the second order. On the other hand, is equivalent to y. But it is not true that the difference of p /3-y' = aS and y', namely, , _ is an infinitesimal of the second order. It is obviously of order 3. Thus replacing y by an equivalent infinitesimal has here changed the order of the difference j8 — y. ^^t yu,c -^^/. ,v^ 3. Differentials. Let y=f(x) be a function of x, and let Dj/ be its derivative : lim^ = i?^. ^^/^i A«=a) Ax ^ : Let the difference Ay /Ax — D^y be denoted by e. Then , Ax and (1) Ay = DjyAx + eAx. Since x is the independent variable, Ax can be taken as the principal infinitesimal. D^y does not vary with Ax; it is a constant, for we are considering its value at a fixed point X = Xa. Since, moreover, D^y is not in general zero, equation (1) represents Ay as the sum of its principal part, DjjAx, and an infinitesimal of higher order, €Ax. Definition of a Differential. The expression D^yAx is called the differential of the function, and is denoted by dy : (2) dy = D,yAx, or df(x)=DJ(x)Ax. (read : " differential y " or " differential f{x) " or " dy," etc.). Thus if y = a?, dy — 2xAx, or dx^ = 2xAx. 92 CALCULUS Since the definition (2) holds for every function y =f(x), it can be applied to the particular function Hence •^('")=^- (3) dx = D^x Ak = Ak. But it is not ia general true that i^y and dy are equal, since t is in general different from 0. Thus we see that the differen- tial of the independent variable is equal to the increment of that variable ; but the differential of the dependent variable is not in general equal to the increment of thai variable. By means of (3) equation (2) can now be written in the form (4) dy = D^dx. Hence (5) dx Geometrically, the increment A?/ of the function is repre- sented by the line MP', Fig. 33 ; and the differential, dy, is equal to MQ, for from (5) " tan T = — dx or dy = dx tan t. In other words. Ay repre- sents the distance from the level of P to the curve, when x = x' ; dy, the dis- tance from the level of P to the tangent. Moreover, the difference Ay — dy = eAx is shown geometrically as the line QP', and is obviously from th& figure an infinitesimal of higher order than Aa; = PM. It is also clear from the figure that Ay and dy are equal when and only when the curve y = /(a;) is a straight line ; i.e. INFINITESIMALS AND DIFFERENTIALS 93 when f(x) is a linear function, f{x) = ax + 6. Hitherto x has been taken as the independent variable, Aa; as the principal infinitesimal. We come now to the theorem on which the whole value of differentials for the purpose of performing differentiation depends. Theoeem. The relation (4) : dy = D,ydx, is true, even when x and y are both dependent on a third vari- able, t. Suppose, namely, that x and y come to us as functions of a third variable, t : (6) a;=<^(0, 2/ = ^(0, and that, when we eliminate t between these two equations, we obtain the function „, ^ Then dx and dy have the following values, in accordance with the above definition, since t, not x, is now the independent variable, At the principal infinitesimal : dy = Diy At, dx = D^x At. We wish to prove that dy = Dj/ dx. Now by Theorem V of Chap. II, § 6 : D,y = Dj/D,x. Hence, multiplying through by At, we get : D,yAt = D^y- DiXAt, or dy = D^dx, q. e.d, 94 CALCULUS With this theorem the explicit use of Theorem V in Chap. II, § 5 disappears, Formula V of that theorem now taking on the form of an algebraic identity : du _ du dy dx dydx To this fact is due the chief advantage of differentials in the technique of differentiation. Differentials of Higher Order. It is possible to introduce differentials of higher order by a similar definition : (7) dH/ = DJh,Aifi, ^y = DJyA3f, etc., X being the independent variable. We should then have by (3) (8) d^=DJh/dx^ or ^,= ^-'^' etc. Unfortunately, however, relation (8) does not contiaue to hold when x and y both depend on a third variable, t. For example, suppose ^„ , ^„ ^ ' ^^ x = t% y = a + t\ Then y = a + x. - When t is taken as the independent variable, we have ac- cording to relation (8) : d^ = nfydt^ = 2dt'^; and since dx = 2tdt, it follows that ^ ^ 2dt' ^ J^ ^ J_ . dx^~4:Pdfi~2fi~2x' On the other hand, when x is taken as the independent vari- able, relation (8) becomes and consequently ^2^^ dh) = D,hfdx^=% d^_ da?' INFINITESIMALS AND DIFFERENTIALS 95 Thus the quotient, — ^, is seen to have two entirely distinct values according as f or a; is taken as the independent variable. We will agree, therefore, to discard this definition. The nota- tion — ^ as meaning DJ'y is, however, universally used in the Calculus, and so we wUl accept the definitions ^=D^% g=2>.V, etc., interpreting the left-hand sides of these equations, however, not as ratios, but as a single, homogeneous (and altogether clumsy !) notation for that which is expressed more simply by Cauchy's D. Remark. The operator D^ shall be written when desired as —■ Thus dx X)j. appears as d X a — X dxa — X Again, the equation appears as d'y_ d dy da? dx dx Finally, the following notation is sometimes used : — = D^y dx, -2 = n sj, ^ etc. dx '" ' da;2 '^ ' \^i. Technique of Differentiation. Consider, for example, Formula II, Chapter II, § 6 : On writing this formula in terms of differentials, we have d{u + v) _ du . dv dx dx dx 96 CALCULUS Now multiply this equation through by dx : d(u + v)=du + dv. Hence the theorem : 7%e differential of the sum of two ftmctions is equal to the sum of the differentials of these functions. The others of the General Formulas, Chapter II, §§ 6, 7, can be treated in a similar way and lead to corresponding theorems in differentials, embodied in the following important group of formulas. Gbnbkal Formulas of Diffkkbntiation. I. d(cu)=cdu. II. d{u + v)= du + dv. III. d{uv)= udv + vdu. -TTT ,M vdu — udv V v'- As already explained, Theorem V reduces to an obvious algebraic identity : w w /7 (a/Hi CEW WV dx dy dx' and so does not need to be tabulated. Of the special formulas hitherto considered, only two need be tabulated, namely : Spboial Formulas of Difpbkentiation. 1. dc = 0. The first of these formulas says that the differential of a constant is zero. The second is valid, not only when x is the independent variable, but when x is any function whatever of the independent variable, t. Thus if (1) M=vr^ INFINITESIMALS AND DIFFERENTIALS 97 and we set (2) x = l-t, equation (1) becomes (3) u = x^. Hence du = -| x~^ dx. But dx = dl + d{—t)=0 — dt, and thus . -dt du 1 du = ^^^— or 2V1 - 1 di 2Vl - 1 The student should copy off neatly on a card the size of a postal the General Formulas I-IV, the Special Formulas 1., 2., leaving room for a few further special formulas. All the difEerentiations of the elementary function of the Calculus are based on these two groups of formulas. To differentiate a function means henceforth to find either its derivative or its differential. Of course, when one of these is known, the other can be found by merely multiplying or dividing by the difEerential of the independent variable. We proceed to show by a few typical examples how differen- tials are used in differentiation. Example 1. Let m = 12 — Sa; + la?. To find du. Take the differential of each side of this equation, and apply at the same time Formula II : du = c?(12)+ d{- 5x)+ d{l3?). By Formula 1, d(12) = 0. By Formula I, d{—5x)=: — 5dx and d(7af^=7da?. Hence du = — 5dx + 21ifidx = (-5 + 21x^)dx and ^ = -6 + 21a;«. dx 98 CALCULUS These steps correspond precisely to the steps the student would take if he were using derivatives, only he would not have written them all out in detail. He would have written down at sight : _ f. oh » He can avail himself of the facility he has already acquired and shorten the work as follows. Since du = Dj/idx, he can begin by writing du=( )dx, and then fill in the parenthesis with the derivative.* Example 2. Let _ a^ — a? To find_dM. By Formula IV we have : ^ ^ (a= + x'^)d(a^ — g') - (g" — x'')dia' + a;') (a2 + a;2)2 ^ (a^ + x-'-)(-2x dx) - (g- - x^)(2xdx) (a2 + a;2)2 ia'^xdx du 4 a^x dx (a2 + xy The student would probably prefer to work this example as follows. Remembering that du = Djjt, dx, * The student must be careful not to omit any differentials. ' If one term of an equation has a differential as a factor, every term must have a differential as a factor. Such an equation as dM = — 5 + 21 x2 is absurd, since the left-hand side is an infinitesimal and the right-hand not. Moreover, there is no such thing as djU. INFINITESIMALS AND DIFFERENTIALS 99 iDegiii by writing du = dx, and then fill in the fraction by the old familiar methods of Chapter II. In the two examples just considered, the processes with difEerentials correspond precisely to those with derivatives^ with which the student is already familiar. , This will always be true in any differentiation in which composite functions are not involved ; i.e. whenever, according to our earlier methods, the vanished Theorem V of Chapter II, § 8 was not used. It is in the diEEerentiation of composite functions that the method of differentials presents advantages over the earlier method. We turn in the next paragraphs to such examples. EXERCISES Differentiate each of the following functions by the method of differentials, and test the result by the methods of Chap- ter II. i^"!. u = x^ — Zx + 1. 2. y ^=a+bx + ex'': 3. w = a' — »'. 4. s = 96<-16«2. 5. s = vat + ^gt\ 1 — x A T —2dx 6. M = - Ans. du = - Ans. du : = 3xMx -,3dx. Ans. dy = = bdx +2cxdx. Ans. dw = — Sz^dz. Ans. ^=96 dt -32t. ■Ans. ^=vo + gt. dt 1 + x (1 + xy '^ Ans. dy^^"-'"'^" " 1+x^ " (l-l-a;2)2 1 + x + x^ . ,a;2 — 1, 8. g= ^ ^ — • .4ms. dz = —- dx. 2x 2x^ 3-2x+a? ,„ n* — ^i 9. u = - , , • 10. y = - 4: + x^ — a? a^ + a'^x^ + X^ 100 CALCULUS 5. Continuation. Differentiation of Composite Fanctions. Example 3. Let u = Vl + x+xK To find — . dx Here, we begin by computing du. To do this, introduce a new variable, y, setting 2/ = 1 + a; + a;'. Then u = y^. Next, take the differential of each side of this equation. By- Special Formula 2 above, du = dy^= \y~^ dy. Moreover, dy = {1 + 2 x)dx. Hence d« = Jl±l^^ 2V1 -\- x+x^ and- du _ 1 + 20! dx 2-Vl+x + x^ Let the student carry through the above differentiation by the methods of Chapter II and compare his work step by step with the foregoing. He will find that, although the two methods are in substance the same," the method of differentials is simpler in form, since no explicit use of Theorem V here is made. Abbreviated Method.* The solution by differentials can be still further abbreviated by not introducing explicitly a new * The student should not hasten to take this step himself. He will do well to omit the text that follows till he has worked a score or more of problems in differentiating composite functions as set forth under Ex- ample 3, introducing each time explicitly a new variable, as y, z, etc. Not until he comes himself to feel that the abbreviation is an aid, should he attempt to use it. INFINITESIMALS AND DIFFERENTIALS 101 variable, y. The problem is to find du, when u=(l + x + a:2)i Now, Special Formula 2, as has already been pointed out, holds, not merely when x is the independent variable, but for any function whatsoever. It might, for example, equally well be written in the form : d [<^(a;)]" = ra[^(a;)]»-id<^(a!). In the present ease, then, the content of that theorem, — the essential and complete truth it contains, — enables us to write down at once the equation : - d{l + x + x^)^= 1(1 + a; -I- a;^)"* d(l + x + x^). This last differential is computed at sight, and thus the answer is obtained in two steps. Even these two steps are carried out mentally as a single process, when the student has reached the highest point in the technique of differentiation. He then thinks of the formula : J r dx d-\/x = - 2y/x realizes that it holds, not merely when x is the independent variable, but for any function of x, and so writes down first the easy part of the right-hand side of the equation, thus : d-\/l + a; + a;2 = 2V1 + a; + a;2' carrying in his head the fact that the numerator is the differen- tial of the radicand, i.e. d(l + x + x^). This differentiation he performs mentally, and thus has the final answer with no intermediate work on paper : 102 CALCULUS Example 4. The method of differentials is especially use- ful in the case of implicit functions. Thus, to find the deriva- tive of y with respect to x when a? — ^xy-\-2y* — l. Take the differential of each side : ^x^dx - 3xdy - 3ydx + 8fdy = 0. Next, collect the terms in dx by themselves ; the others will contain dy as a factor : (3x^ - 3y)dx + (8f - 3x)dy = 0. Hence ^^3_y-3^, dx sy' — 3a! EXERCISES Differentiate the following twelve functions by the method of differentials and also by the methods of Chapter II (in either order), introducing each time explicitly the auxiliary variable, if one is used. 1/ 1. u=V a^ + a^^ + a^. Ans. du = J^^±^^^- Va* -f- aV + !B* T 3- y = — • Ans. dy = 3. M = Ans. du = ■ dx J " 3. u = - j3.ns. au = — -• T -y 1-x (l-®)' A If Suggestion. Introduce an auxiliary variable y = l — x. \ Then u = i /V 1^ 4. u=— — - Ans. — - \ -, " ' (1 - xy dx (1 - x)' I v^ , 1 . dy — 2x v/ ■^ H-as" dx (l + x^y INFINITESIMALS AND DIFFERENTIALS 103 6. s = — • Ans. — = — - (a + ty y dt (a 4 ty 7. 2x''-xy + iy' = 5. Ans. ^ = ^'"~^ . dx x — Sy 8. a!2/ = al ^ns. ^ = -1. dx X 9. 2/2 = 2ma;. Ans. ^=^. da; 2/ 10. ^ + ^=L ^ns. ^ = -^. 11. 2a!2 + 3y2 = 10. ^ns. ^ = _?^. da; 3y 12. 2a!2/-a; + !/ = 0. Ans. ^ = 3lJi1|. da 2a! 4-1 The student can work the problems at the end of Chapter II by the method of differentials. Tor further practice, if de- sired, the following examples are appended. du la^—lx"- — ! 13. u={x^+l)y/3?-x. Ans. — = dx 2 Va;' ■ • X 14. y={x + 2V)(x-hY. Ans. ^ = 3(x^-b^). dx 15. u = - Va^ - x^ 16. u = yj: la — X <l— 17. u=- X X — a Ar, ,.s. *i= dx a2 V(a^- xy Ans. du_ dx a 2x^ax- — x' Ans. du d? ^2ax — x^ dx V(2aa; — 0:2)3 18. u = f^l±^\ ■ Ans. *f=2- »^ L^^Y a; / da; a;* — a^ 19. ,=(t±b^y. Ans. ^=42^i^^ \ y^ J dy 2/5 104 CALCULUS 20. u = — — Va" — a?. Ans. — ■= ■ a? cfa; af Va'- — a? „, ja^— x-\-l . du 21. w=-\/ ■ — • Ans. — = ^x'' + x + l dx 3a* + a; + 1 dx (3-2+ X +. 1) Va^ + x^ + l 23. u __ (x — a^)' . du _ S-Vx — x^ ~ x^ ' dx 3a^ = (xi-aiy. Ans. *^ = iMzi^. dx 3a,f du 24. u = x(x^ + 5)\ Ans. ^ = 5(»' + l)(a!3 + 5)i dx 25. a;' + y^ = a'. -4ns. -^ = — \;'" . da; *a; CHAPTER V TRIGONOMETRIC FUNCTIONS 1. Radian Measure. In Trigonometry, the radian measure of an angle was introduced, apparently for no good purpose. The reason lies in the importance for the Calculus of this new system of measurement, and will become clear in the next paragraph, when we come to differentiate the sine. We will first recall the definition. Let a circle be described with its centre at the vertex O of the angle ; let r denote the length of the radius of the circle and s, that of the intercepted arc. Then the radian measure, 6, of the angle is defined as the ratio s/r : (1) For a right angle, s = — - , and hence 6 ■■ Fig. 34 A straight angle has the measure 61 = T = 3.14159 26535 89793 • - .. Let <l> be the measure of the given angle in degrees. Then d and <j> are proportional, e = c<J3, where c is a constant. To determine c, use a convenient angle whose measure is known in both systems ; for example, a straight angle. For the latter, and 105 <l> = 180. 106 CALCULUS Substitating these values in the above equation we find : and hence 7r = cl80, ''"180' (2) 180 '^' TT This equation can also be written in the form (3) TT 180 and thus an easUy remembered rule of conversion from radian measure to degree measure, or the opposite, obtained: The radian measure of an angle is to w as its degree measure is to 180. The unit of angle in radian measure, i.e. the angle for which ^ = 1 and hence s= r, is called the radian. It is obvious geometrically that it is a little less than 60°. Its precise value (to hundredths of a sec- ond) is given by (2) : <!> \^, = ^ = 57° 17' 44.81" (= 57.29578°). TT On the other hand, the radian measure of an angle of 1° is 1 = J!L = .01745 32925 19943 -. '*-' 180 The student should practice expressing the more important angles, as 30°, 46°, 60°, 90°, 120°, etc., in radian measure until he is thoroughly familiar with the new representation for them. If, in particular, the radius of the circle is taken as Tinity, then 6 and s are the same number : (4) 6 = s, when r = 1 ; or the arc is equal to the angle. Thus the radian measure of an angle might have been defined as the length of the intercepted TRIGONOMETRIC FUNCTIONS 107 are in the unit circle (i.e. the circle of unit radius with its centre at 0). Graph of sin x. It is important for the student to make an accurately drawn graph of the function y = siax, X being taken in radian measure. Let the unit of length, as usual, be the same on both axes, and let it be chosen as 1 cm. For this purpose Peirce's Table of Integrals (the table of Trigonometric Functions near the end) is especially convenient, since the outside column gives the angles in radian measure, and thus as many points of the graph as are desired can be plotted directly from the tables. j/=sin X Fig. 35 Since sin (tt — a;) = sin x each determination of the coordinates {x, y) of a point on the graph, for which < a; < ^ yields at once a second point, namely (tt — x, y). Thus one arch of the curve is readily con- structed from the Tables.* From this arch a templet, or curved ruler, is made as fol- lows. Lay a card under the arch and with a needle prick through enough points so that the templet can be cut ac- curately with the scissors. By means of the templet further arches can be drawn me- chanically, and thus the curve is readily continued in both * The graph could be made directly without tables from purely geomet- rical considerations. Draw a circle of unit radius. Construct geomet^ rically convenient angles, as those obtained from a right angle by successive bisectors. Measure any one of these angles, X ABP„, in ra^ dians and this number will be the abscissa of the point on the graph, the 108 CALCULUS directions to the edges of the paper.* Put this curve in the upper quarter of a sheet of centimetre paper. The graph brings out clearly the property of the function expressed by the word periodic. The function admits the period 2ir, since sm (a; + 2 tt) = sin a; Graph of cos x. By means of the templet the graph of the function y = cos X can now be drawn mechanically. This function also admits the period 27r : / „ ^ cos (a; + 2 tt) = cos x. ordinate being the perpendicular dropped from P„ on tlie line BA. Thus, if n = 3, the coordinates of tlie point on the graph are : = ^ = 1.18, y = .92. A second point of the arch, that corresponding to Pj, has the same y, its coordinate be- ing x = 7r-^=1.96, y=.92. 8 Of course, the distance ir must be laid off on the axis of x by measui'e- ment ; it cannot be constructed geometrically from the unit length. This done, the further abscissae are found by successive bisectors. * In order to obtain the most satisfactory figure, observe that the curve has a point of inflection at each of its intersections with the axis of X, the tangent there making an angle of ± 45° with that axis. Since a curve separates very slowly from an inflectional tangent, it will be well to draw these tangents vrith a ruler. On laying dovro the templet, the curve can then be ruled in from the latter vrith great accuracy. It will not separate sensibly from its tangent for a considerable distance from a point of inflection. TRIGONOMETRIC FUNCTIONS 109 Put the graph in the second quarter of the sheet, choosing the axis of y for this curve in the same vertical line as the axis of y for the sine curve above. There remains the lower half of the sheet for the next graph. Graph of tan x. The same tables make it easy to plot points profusely on the graph of the function y = tan x in the interval 0^ a: < -. Take the axis of y in the same ver- i I I I 2/=tan± ¥ ^ Fig. 38 tical line as in the case of the preceding graphs. This done, a second templet is made and by means of it the graph is drawn mechanically for values of x such that — ^ < a; < 0. It is desirable furthermore to plot the function in the two adjacent intervals no CALCULUS "■ ^ « ^ Stt -T<^<-2' in order to suggest the fact that this function admits the period it: , , ^ . tan {x + ir) — tan x. 2. Differentiation of sin x. To differentiate the function (1) y = smx, apply the definition of a derivative given iu Chap. II, § 1. Give to X an arbitrary value cBb and compute the corre- sponding value y^oiy; yo = sin Xo- Then give x an increment Ax, and compute again the corre- sponding value of y : yo + Ay = sia (xa + Ax). Hence Ay = sin (ajo + Ax) — sin Xq, Fig. 39 M' M (2) Ay _ sin (xp + Ax) — sin a;,. Ax Ax It is at this poiat in the process that the specific properties of the function sin a; come into play. Here, the representar tion of sin a; by means of the unit circle, familiar from the beginning of Trigonometry, is the key to the solution. From the figure it is clear that siu Xo = MP, sia {x^ + Ax) = M'P', Ay = sin (xq + Ax) — sin aio = QP', Ax = PP'. Hence (3) Ax ppi' TRIGONOMETRIC FUNCTIONS 111 and so we want to know the limit approached by the latter limS^. By virtue of the Fundamental Theorem of Chap. IV, § 2, we can replace this ratio by a simpler one, since the arc PP and the chord PP' are equivalent infinitesimals : * lim5 = l. PP Hence lim ^' = lim ^. p'=p jypf F'=p ppt On the other hand, the triangle QPP' is a triangle of refer- ence for the 2^ QPP = <f>, and so ■ ■ ^ — = sin <A. PP When P approaches P, the secant PP' (i.e. the indefinite line determined by the two points P and P) approaches the tan- gent PT at P, and thus lim <i> = -4. qPT =1-01^ p'=p J Finally, then, lim ^ — = lim sin <^ = sinf ^ — Xq )= cos Xq, P'=p PP p'=p \2 J and consequently j- ^^ A^ _ ^^^ ^^^ A»a) Aa; * The student should assure himself of the truth of this statement by visualizing the figure (making an accurate drawing with ruler and com- pass for angles of 30°, 15°, and 7J°, the circle used being 10 in. in diameter) and realizing that, when P' is near P, the difference in length between the arc and the chord is but a minute per cent of the length of either one. A formal proof will be found below. 112 CALCULUS or, on dropping the subscript, (4) DjSin X = cos x. This theorem gives rise to the following theorem in dif- ferentials : (5) d sin X = cos x dx. Reason for the Radian. The reason for measuring angles in terms of the radian as the unit now becomes clear. Had we used the degree, the increment \x would not have been equal to PP ; we should have had : Aa; PP' . 180 ^^, = , or Aa; = PP\ 360 2 TT IT Hence (3) would have read : Ay_ TT QP Ax 180 ppi' and thus the formula of differentiation would have become : D^ sin X = -^ cos x. 180 The saving of labor in not being obliged to multiply by this constant each time we differentiate is great. Still more impor- tant, however, is the elimination of a multiplier which is of the nature of an extraneous constant, whose presence would have obscured the essential simplicity of the formulas of the Calculus. EXERCISE Prove in a similar manner that D^ cos x = — sin x. 3. Certain Limits. In the foregoing paragraph we have made use of the fact that the ratio of the arc to the chord approaches 1 as its limit. A. formal proof of this theorem, based on the TRIGONOMETRIC FUNCTIONS 113 axioms of geometry, can be given as follows. Draw the tan- gent at P and erect a perpendicular at P' cutting the tangent iu Q. Denote the angle 2(! P'PQ by a. Then . ^ PP <PP' <Pq + FQ; P for i) a straight line is the shortest dis- tance between two points ; and ii) a convex curved line is less than a convex broken line which envelops it and has the same extremities. But PQ = EJ^, P'Q = PPta.Tia. cos a Hence J!P' 1 1 < < 1- tan a. PP cos a When « approaches 0, the right-hand member of the double inequality approaches 1 ; hence the middle member must also approach 1, or -^ lim = 1, q. e. d. PP The foregoing proof holds, not merely for a circle, but for any curve with a convex arc PP'. Consequently the theorem is established generally. The Limit lim ^^^. From Pig. 41 ^"^ it is clear that - MA JfP=sin«, AP=a, Fig. 41 and hence sin a _ MP « " AP By direct inspection of the figure it is seen, then, that .(1) lim?i^=l. 114 CALCULUS A formal proof of this equation can be given as foUowa From Fig. 42 ^ PP' = 2sina, PP' = 2a. Hence sina_PP' and therefore, by the proposition just established, lim?^ = lim^'=l. 0=^0 a p'=pppi Another Proof of (1). The area of the sector OAP, Fig. 42, is \ a, and it obviously lies between the areas of the triangles OMP and OPN. Hence ^ sin a. cos « < ^ a < ^ tan a or cos a < -^— < sm a cos a •jpr When « approaches 0, each of the ex- treme terms approaches 1, and so the middle term must also do so, q. e. d. From Peirce's Tables, p. 130, we see that sin 4° 40' = .0814, and the same angle, measured in radians, also has the value .0814, to three significant figures. Thus for values of a not exceeding .0814, sin a differs from a by less than one part in 800, or one-eighth of one per cent. The Limits lim ^ - "os « ^^^ lim ^~'^°^" - From Fig. 42, the first of these limits is seen to have the value : lim- cos« = 0. TRIGONOMETRIC FUNCTIONS 115 A formal proof can be derived at once by the method em- ployed in the evaluation of the next limit, 1 lim- cos a Expressing 1 — cos a. in terms of the half angle, we have l-cos« = 2sin2^. Hence and 1 — cos a Ssin^? 2 1 ''2 lim l-cos« _li,^ «2 2 -^ . Of a 2 . EXERCISES In the accompanying figure determine the following limits when ot approaches : 1. lim 2. lim 3. lim AR MP AQ AP BQ MP Ans. — 2 Ans. 1. 4. lim FlQ. 43 6. lim — — • 7. lim HP AP PQ AN 6. lim 8. lim PN AP BQ pn' Determiae the principal part of each of the following infini- tesimals, referred to a as principal infinitesimal : 9. MP. Ans. a. 10. PB. Ans.^a. 11. BQ. 12. PN. 13. AQ. 15. PQ. 16. MN. 14. MA. Ans. I a'. 17. AQ-MP. 116 CALCULUS 4. Critique of the Poregoiag Differentiation. The differenti- ation of sin a; as given in § 1 has the advantage of being direct and lucid, and thus easily remembered. Each analytic step is mirrored in a simple geometric construction. It has the dis- advantage, however, of incompleteness. For, first, we have allowed Ax, in approaching 0, to pass only through positive values ; and secondly we have assumed Xg to lie between and ^ IT. Hence there are in all seven more cases to consider. An analytic method that is simple and at the same time general is the following. Eecall the Addition Theorem for the sine : sin (a +b)= sin a cos b + cos a sin b, sin (a — 6) = sin a cos 6 — cos a sin 6, whence sin (a -(- 6) — sin (a — &) = 2 cos a sin h. Let a + b = XQ+ Ax, a — b = XQ. , Solving these last equations for a and b, we get : , Ax , Ax and the difference-quotient becomes . Ax sm — Aw / Ax\ 2 -=cos(^a.o+-^j-^- 2 The first factor on the right approaches the limit cossjo when Aa; approaches 0. On setting ^Ax= a, the second fac- tor becomes sma Hence the factor approaches 1. Thus lim — = cos Xq, ctx^Ax TRIGONOMETRIC FUNCTIONSv 117 or, on dropping the subscript, Z)j. sin X = cos X. 5. Differentiation of eosa;, tanw, etc. To differentiate the function cos x, introduce a new variable, y, by the equation • y=^—x. Hence x = - — y, and cos x = cos ( — - y]= sin y. Taking the differential of each side of the equation thus ob- tained, we have : dcosa; = dsmy = oos ydy. But cosy = sin X and dy = -dx. Hence 1 (1) d cos aj = — sin a; da;. To differentiate the function tan x, set .„„ sin a; tan x = cos a; Hence dtau X cosa;dsina; — sina;d COS^ X cos a; _ cos" a; da; 4- sin^ a;da; _ dx and thus cos^a cos^a;' (2) dtana;— sec^xda;. It is shown in a similar manner (or by setting x= — — y in the equation just deduced) that (3) d cot a; = — cs c^ a; da;. These are the important formulas of differentiation for T;he trigonometric functions. By means of them all other differen- tiations of these functions can be readily performed. Thus, 118 CALCULUS to difEerentiate the function sec x, set sec X = (cos a;)~*- Then dsecx = = . cos^ X cos^ X It is not desirable to tabulate the result, since one rarely has occasion to difEerentiate either sec a; or csco;, and when 'the occasion does arise, the difEerentiation can be worked out directly, as above. The student should now add to his card of Special Formulas the four main formulas just obtained. This card will now read as follows : 1. dc = 0. 2. - dx" = ma;"~i dx. 3. d sin X = cos x dx. 4. d cos x = — sin x dx. 5. . d tan x = sec" x dx. 6. d cot x = — esc" X dx. 6. Shop Work. To acquire facility in the use of the new results, the student should work a generous number of simple examples, for which the following are typical. Example 1. To differentiate the function M= sin aa;. Let y = ax. Then M = sin y. and du = daiay = cosydy. But dy = a dx. Hence, substituting, we have du = acoaaxdx or — sin aa; = a cos ax. dx TRIGONOMETRIC FUNCTIONS 119 The solution can be abbreviated as follows. The equation d sin X = cos x dx is true, not merely when x is the independent variable. It holds, for example, in the form dsmy = COS ydy, where y is any function of x. Hence we can write immediately dsin ax = cos ax d{ax), and thus obtain the result d sin ax^a cos ax dx. Example 2. To differentiate the function M=V1 — fc^sin",^. Let z = 1 — fc2 sin" <^. Then u = z^ du = dz' = ^z~^dz; dz = -K'dairs?^. Let y = sin <j}. Then dy = cos tj> d^ and dsin' </i = dy^ = 23/% = 2sin<^ cos <^d<^. Hence c?m=^«"^(— 2fc^sin<^ cos (^d<^) du fc2 sin d cos A or — = ^ -!— . d<l> Vl — A;2 sin2 </. Example 3. If siax + 3my = x- y, to find -^. Take the differential of each side of the equation: dx cos xdx + cos ydy = dx — dy. Hence (cos x — V)dx + (cos y + l)dy = , dy _ 1 — cos X dx 1 + cos 3^ 120 CALCULUS EXERCISES Differentiate the following functions. 1. M = cos oa;. 2. y = cos^x. 3. y = CSC X. X 4. M = tan— ■ 2 5. M = cot2a;. 6. M = sec3a;. 8. u = sin' X. 10. u = x + tan X. 12. M = sec^ X. sin a; 14. w = - 1 — cos X 15. u = Vl + cos X. 1 — cos a; 16. u = 18. M = 1 + cos a; sin a; 20. u = a + b cos X 1 sin a; + cos x 22.* M = vers X. 23.* It = covers x. * The versed sine and the coversed siTie are defined as follows : vers a; = 1 — cos a; ; covers a; = 1 — sin a. dx — a sin ax. dx — 2 sin a; cos a;. dy _ dx esc' x cos X. du _ dx 2 2 du_ -2csc2 2a!. 7, . M = tan^ ax. 9 . U = 1 — sin X. 11 . u = cos' X. 13 . u = sin X cos X. du_ dx _icsc25. 2- 2 du_ dx : 1 sinf. a/2 2 17 . u = 1 + sin a; 1 — sin a; 19 . u = 1 a cos a; + 6 sin a; 21 . u = 1 (a + b cos a;)^ du dx~ : sin X. du_ dx — cos a;. TRIGONOMETRIC FUNCTIONS 121 cos? 24. u = x sin. 2x. 25. m = X 28. M = taii-^- 29. u = 5iBJ!^. 1 -^ X X 30. M = sin a; + cos 2 a;. 31. m = a;^ cos tts;. 1 «„ cos d> 32. M=- 33. M = — z=:^===- Vl-A!''sin2^ Vl — &2 sini* <^ ■ / . \ (^V COS (a; + w) — cos 11 34. a!COS«= sm(!B + w). ^^ = -, '- ^ "^ r-^- dx sm (a; + y) + a; sm y 35. tana;— cot 2/= sin a; sin 3/. 36. sin x + sin jr = 1. 37. tan + tan <^ = 2 tan <^ tan 6. 38. a; = 2/ sin y. 7. Maxima and Minima. By means of the new functions studied in this chapter the range of problems in maxima and minima which can be treated by the Calculus has been materi- ally enlarged. No new principles are involved ; the student should go over carefully the paragraphs of Chap. Ill relating to this subject, before he proceeds farther with the present paragraph. Example 1. A man in a rowboat 1 mile off shore wishes to go to a point which is 2 miles inland and 4 miles up the beach. If he can row at the rate of 5 miles an hour, but can walk only 3 miles an hour after he lands, in what direction should he row in order to get to his destination in the shortest possible time ? In the first place, it is clear that the straight line AEB is , not the best path. For, if he rows toward a point P slightly farther up the beach, the amount by which he lengthens the leg AP of his path is very nearly equal to the amount by which 122 CALCULUS he shortens the leg PB.* Consequently the time is short- ened. On the other hand, P obviously ought not to be taken so far up the beach as D. The minimum occurs, therefore, for some in- termediate point. Let the angles 0, <f> be taken as indicated in the figure. Then, since i = -, Fig. 44 V time from A to P = ^^^ = — - — ; 5 6 cos 6 time from P to 5=:^= ^ ^'7 X/ B 2 X sy V '' ^-<r P D 1 A 3 cos (9 Hence the total time, u, which is to be made a minimum is 1 . 2 (1) ; + ; 5 cos 6 3cos ^ Moreover, 6 and <^ are connected with each other by a rela- tion which is readily obtained by expressing the distance CD in two ways : (2) tane +2tan<^ = 4. We are now ready to compute du/d6 and set it equal to : J _ _ dcos^ _ 2 d cos <i> ~ 5 eos^ 3 eos2 ^ (3) sec^ sin 6 ■,„ , 2 sec'' tb sin <A , . = g (w-l ^ ^dcj}-, du _ sec' 6 sin 2 sec' <f> sin <^ d<j) d6~ 5 3 W * Let the student not leave this statement till he is absolutely con- vinced of its truth. An accurate figure on a large scale will bring the fact out clearly. TRIGONOMETRIC FUNCTIONS 123 On setting du/dO = 0, we obtain the equation : ,.-. setfi 6 ajg 6 _ _ 2 sec' <^ sin <^ d^ 5 Z dd Next, differentiate (2) : sec'= Odd + 2 Bec2 ^ d</, = 0, (5) sec25l = -2sec2^^. Now, divide equation (4) by equation (5) : * ,a\ sin 6 gin <A sin 6 5 (o) • =i^ — 2: or = - • ^ '. 5/3 sinc^ 3 The result, stated in words, is as follows : sia 6 is to Sia <^ as the velocity in water is to the velocity on land. Let the student work the general . problem, in which all the data are take^ in literal form, and verify the general result just stated. In order actually to determine 9, equations (2) and (6) must be solved as simultaneous : ,„^ I tan 6 + 2 tan <^ = 4, ^ ' . l3sine = 5sin<^. This is done best by the method of Trial and Error, as it is called in Physics ; Successive Approximations being the name usually given to it in Mathematics. It is a most important method in both sciences, and the student should let no oppor- tunity go by to use the method whenever, as here, he meets a case which calls for it. Cf. Chap. VII, § 5. The Corresponding Problem in Optics. We have stated and solved a problem which is not lacking in interest, but which appears to have no scientiiic importance. This very problem, however, occurs in Optics. The velocity of light is different id * i.e. divide the left-hand side of (4) by the left-hand side of (5) for a new left-hand side ; and do the same thing for the right-hand sides. 124 CALCULUS different media, such as air and water. Suppose two media to be in contact with each other, the common boundary being a plane. Let ^ be a luminous point, from which rays emanate in all directions. When the rays strike the bounding surface, they are all refracted and enter the second medium in case the velocity of light in that medium is less than in the first. One of the refracted rays will pass throttgh a given point B. And now the law of light is that the time required for the light to pass from ^ to B is less for this path than for any other possible path. If, then, the velocity of light in the first medium is m * and in the second medium, v, we have : sin 6 u sin<^ V where n is the iTidex of refraction for the passage from the first medium into the second. EXERCISES 1. A wall 27 ft. high is 64 ft. from a house. Find the length of the shortest ladder that will reach the house if one end rests on the ground outside the wall. Take the angle which the ladder makes with the horizontal as the independent variable. 2. The equal sides of an isosceles triangles are each 8 in. long, the base being variable. Show that the triangle of maximum area is the one which has a right angle. Take one of the base angles as the independent variable, ^. 3. A gutter is to be made out of a long strip of copper 9 in. wide by bending the strip along two lines parallel to the edges and distant respectively 3 in. from an edge. Thus the cross-section will be a broken line, made up of three straight lines, each 3 in. long. How wide should the gutter be a;t the * The letter u used here has nothing to do with the v, used above in solving the problem. TRIGONOMETRIC FUNCTIONS 125 top, in order that its carrying capacity may be as great as possible ? Ans. 6 ia. 4. Johnny is to have a piece of pie, the perimeter of which is to be 12 in. If Johnny may choose the plate on which the pie is to be baked, what size plate would he naturally select ? 5. A can-buoy in the form of a double cone is to be made from two equal circular iron plates by cutting out a sector from each plate and bending up the plate. If the radius of each plate is a, find the radius of the base of the cone when the buoy is as large as possible. Ans. aVf. 6. From a circular piece of filter paper a sector is to be cut and then bent into the form of a cone of revolution. Show that the largest cone will be obtained if the angle of the sector is .8165 of four right angles. 7. Two solid spheres, whose diameters are 8 in. and 18 in., have their centres 35 in. apart. At what point in their line of centres and between the spheres should a light be placed in order to illuminate the largest amount of spherical surface ? Ans. 8 in. from the centre of the smaller sphere. 8. Find the most economical proportions for a conical tent. 9. A block of stone is to be drawn along the fioor by a rope. Find the angle which the rope should make with the horizontal in order that the tension may be as small as possible. Ans. The angle of friction. 10. A block of stone is to be drawn up an inclined plane by a rope. Find the angle which the rope should make with the plane, in order that the tension in the rope be as small as possible. 11. A statue ten feet high stands on a pedestal that is 50 ft. high. How far ought a man whose eyes are 5 ft. above the ground to stand from the pedestal in order that the statue may subtend the greatest possible angle ? 12. A steel girder 25 ft. long is moved on rollers along a passageway 12.8 ft. wide, and into a corridor at right angles 126 CALCULUS to the passageway. Neglecting the horizontal width of the girder, find how wide the corridor must be in order that the girder may go round the corner. Ans. 5.4 ft. 13. A gutter whose cross-section is an arc of a circle is to be made by bending into shape a strip of copper. If the width of the strip is a, find the radius of the cross-section when the carrying capacity of the gutter is a maximum. Ans. a/ir. 14. A long strip of paper 8 in. wide is cut off square at one end. A corner of this end is folded over on to the opposite side, thus forming a triangle. Find the area of the smallest triangle that can thus be formed. 15. In the preceding question, when will the length of the crease be a minimum ? 16. The captain of a man-of-war saw, one dark night, a privateersman crossing his path at right angles and at a distance ahead of c miles. The privateersman was making a miles an hour, while the man-of-war could make only b miles in the same time. The captain's only hope was to cross the track of the privateersman at as short a distance as possible under his stem, and to disable him by one or two well-directed shots ; so the ship's lights were put out and her course altered in accordance with this plan. Show that the man-of-war crossed the privateersman's track - ■Va!' — b^ miles astern of the latter. If a = 6, this result is absurd. Explain. 17. The illumination of a small plane surface by a luminous point is proportional to the cosine of the angle between the rays of light and the normal to the surface, and inversely pro- portional to the square of the distance of the luminous point from the surface. At what height on the wall should an arc light be placed in order to light most brightly a portion of the floor a ft. distant from the wall ? Ans. About ^^ a ft. above the floor. TRIGONOMETRIC FUNCTIONS 127 18. A town A situated on a straight river, and another town B, a miles farther down the river and 6 miles back from the river, are to be supplied with water from the river pumped by a single station. The main from the waterworks to A will cost $ m per mile and the main to B will cost % n per mile. Where on the river-bank ought the pumps to be placed ? 19. A telegraph pole 25 ft. high is to be braced by a stay 20 ft. long, one end of the stay being fastened to the pole and the other end to a short stake driven into the ground. How far from the pole should the stake be located, ia order that the stay be most effective ? 20. Into a full conical wine-glass whose depth is a and generating angle « there is carefully dropped a spherical ball of such a size as to cause the greatest overflow. Show that the radius of the ball is a.sm a sin a + cos 2 a 21. A foot-ball iield 2 a ft. long and 2 b ft. broad is to be surrounded by a runnhig track consisting of two straight sides (parallel to the length of the field) joined by semicircular ends. The track is to be 4 c ft. long. Show how it should be made ia order that the shortest distance between the track and the foot-ball field may be as great as possible. 22.* The number of ems (or the number of sq. cms. of text) on this page and the breadths of the margius being given, what ought the length and breadth of the page to be that the amount of paper used may be as small as possible ? 23. Assuming that the values of diamonds are proportional, other things being equal, to the squares of their weights, and that a certain diamond which weighs one carat is worth $ m, show that it is safe to pay at least $ 8 m for two diamonds which together weigh 4 carats, if they are of the same quality as the one mentioned. * Exs. 22-25 do not involve Trigonometry. 128 CALCULUS 24. When a voltaic battery of given electromotive force (E volts) and given internal resistance (r ohms) is used to send a steady current through an external circuit of R ohms resistance, an amoimt of work, W, equivalent to ^^ X 10' ergs is done each second in the outside circuit. Show that, if dif- ferent values be given to M, W will be a maximum when E = r. 25. An ice cream cone is to hold one-eighth of a pint. The slant height is I, and half the angle at the vertex is x. Find the value of x that will make the cost of manufacture of the cone a minimum. (Ans. x = 35°.27.) 8. Tangents in Polar Coordinates. Let be the equation of a curve in polar coordinates. We wish to find the direction of its tangent. The direction will be known if we can determine the angle tj/ between the radius vector pro- duced and the tangent. Let P, with the coordinates (tq, 6^, be an arbitrary point iT of the curve and P':(ro-FAr, flo+Afl) a neighboring point. Draw the chord PP and denote the Z OP'P by f . Then obviously lim^'= ^Q. To determine ^qj Pjq 45 drop a perpendic- ular PM from P on the radius vector OP' and draw an arc PN of a circle with as centre. The right triangle MP'P is a triangle of refer- TRIGONOMETRIC FUNCTIONS 129 ence for the angle ij/' and ^ MP Hence P'M (1) cot ^0 = li™- cot ij/' = lim • In the latter ratio we can replace P'M and MP by more convenient infinitesimals ; cf. Chap. IV, § 2. We observe that MP = Tq sin A6 ; hence lim = lim ; — = 1 ; i.e. MP and r^AB are equivalent infinitesimals. Furthermore, P'Jf and P'iV=Ar are also equivalent infini- tesimals. For _,,^ „,,_ , „,^ P'M=P'N+ NM and -?df =?■(, — ro cos Afl. 1 — cos Afl Hence Now, by § 3, NM ■'o- Ad Ar Ar AS limi^ - cos Afl 0. A61 On the other hand, ,• Ar _ j-. and this quantity is not, in general, 0. Hence lim^=0. A8^ Ar Returning to equation (1) we can now write the last limit in the form: p,^ ^ ^ hm = lim = — Dgr; p'=p MP Ae^roAe r^ or, dropping subscripts, (2) coti/r = i2),r. 130 CALCULUS In terms of differentials, this result can be written in either of the two forms : (3) cot^= dr tani/f = rdO rde' ' ^ dr Examvple. Consider the parabola in polar form TO 1 — cos <^ To determine i/». Here, Hence C0tl/f = *"" (1-C0S^)2" m sin <^ d<f> 1 — cos <^ (1 — COS <^)2 md<^ _ sin<^ 1 — COS ^ In particular, at the extremity of the latus rectum, we have : and thus we obtain anew the result that the tangent there makes an angle of 45° with the axis of the parabola. Again, at the vertex, COtU=„=0, ,/r = |, and the tangent there is verified as perpendicular to the axis. From the above equation, cot^ = ?^5^-, ^ 1 — cos ^ a simple relation between ^ and <f> can be deduced. Since sin^ 2 sin* cos* 2 2,,^ ^ — J — ; — =cot5, l-cos</, ,„.„,,^ 2 2sin2 2 2 TRIGONOMETRIC FUNCTIONS 131 it follows that , , .A eot^ = — cot^- But, for any angle, x, cot (w — x)= — cot X. Setting a; = ^ in the above equation, we have : cot (ir — i/f) = cot ^ • Hence * , <4 or, fke supplement of ^ is equal to *. Thus we have a new proof of the familiar property of the parabola, that the tangent at any point P of the curve bisects the angle between the focal radius, OP, and a parallel to the axis, drawn through P. EXERCISES 1. Plot the spiral, r=6, and show that the angle at which it crosses the prime direction whene = 27r is 80° 67'. 2. Plot the spiral, _ 1 Show that it has an asymptote parallel to the prime vector. Suggestion. Consider the distance of a point P of the curve from the prime direction, and find the limit of this distance when 6 approaches 0. Determine the angle at which the radius vector correspond- ing to 6 = 17/2 meets this curve. 3. Plot the cardioid, r = a (1 — cos </>), * The trigonometric equation admits a second solution, namely, (^ _ ^) + T = <p/2. If, however, we agree to take and ^ so that ■^ < 2 TT and '^ ^ < tt, this second solution is ruled out. 132 CALCULUS and show that , , sin d> cot d/ = — — • 1 — cos <l> At what angle is the curve cut by a line through the cusp perpendicular to the axis ? 4. Prove that, for the cardioid, ^ 2 5. Show that the tangent to the cardioid is parallel to the axis of the curve when <^ = |^7r. 6. At what points of the cardioid is the tangent perpen- dicular to the axis of the curve ? 7. Determine the rectangle which circumscribes the cardioid and has two of its sides parallel to the axis of the curve. 8. Show that, for the lemniscate, ?-2 = a2cos2e, the angle i^ is given by the equation : cot i/' = — tan 26. Hence, show that , tt , o /i 'I' = 2 9. At what points of the lemniscate is the tangent parallel to the axis * of the curve ? Ans. At the point for which = 7r/6, and the points which correspond to it by symmetry. 10. The points of the curve at which the tangent, is parallel to the prime vector, are evi- dently those for which •' y = r sm </), * The axis of any curve is a line of symmetry. The lemniscate has two such lines. The axis referred to in the text is that one of these lines which passes through the vertices of the curve. TRIGONOMETRIC FUNCTIONS * 133 considered as a function of <^ through the mediation of the equation of the curve, has a maximum, a minimum, or a certain point of inflection. For these points, then, ^ = r cos d + sin d. — = 0. Show that this condition is equivalent to the one used above in the special cases considered, namely : }fl -\- <^== IT. 11. Plot the curve, r = acos2^, taking a= 5 cm. Show that for this curve cot./' = — 2 tan 26. 12. At what points of the curve of question 11 is the tan- gent parallel to the axis ? Ans. For one of the points, tan 6 = V5 13. Plot the curve, /• = a cos 3fl, taking o = 5 cm. Show that eot.^ = — 3tan3e. 14. At what points of the curve of question 13 is the tan- gent parallel to the axis of the lobe ? Ans. For one of these points, tan =-v/l -| -• 15. The equation _ m 1 4- sin (/> represents a parabola referred to its focus as pole. Give a direct proof that the tangent to this curve at any point bisects the angle formed by the focal radius drawn to this point and a parallel to the axis through the point. 16. Show that the tangent to the hyperbola m r = = 1 — V3 cos <^ 134 ' CALCULUS at the extremity of the latus rectum makes an angle of 60° with the transverse axis. 17. Prove that the -tangent to the ellipse r=^ , V3 — cos ij) at the extremity of the latus rectum makes an angle of 30° •with the major axis. 9. Differential of Arc. Let (1) y =/{<>') be the equation of a given curve. Let P, with the coordinates (x, y),he a variable point, and A a fixed point of the curve. Denote the length of the arc AF by s. Then s is a function of X ; for, when x is given, we know P and thus s. It is possible to determine the derivative of s, D^, as fol- lows. By the Pythagorean Theorem we have (Chap, IV, Fig. 33), PP^ = Aa!2 + ^y^'. Let A* approach as its limit. Then lim f^7= 1 + 1™ (¥)^= 1 +(Ojyy. Since by § 3 the chord PP' and the arc PP'= As are equivalent infinitesimals, it follows from the Fundame ntal Theorem of Chap. IV, § 2 that, in the above equation, PP' can be replaced by As. Hence limf^Y=limf^Y = (i).,)S and consequently (2) (D^sy=i+ip^yy. TRIGONOMETRIC FUNCTIONS 135 On replacing the derivatiTes in (2) by their values in terms of differentials, we have \aaij \dxj or (3) ds^ = dx^ + dy^. This formula is easily interpreted geometrically by means of the triangle PMq, Fig. 33. Since PM= dx and MQ = dy, it follows from the Pythagorean Theorem that (4) PQ = ds. It is obvious geometrically that ds and As differ from each other by an iniiiiitesimal of higher order ; i.e. that they are equivalent infinitesimals.* Formulas for sin t, cos t. From the triangle PMQ we can write down two further formulas : /e\ • dy dx (5) sm T = -^ , cos T = — ds ds These formulas presuppose a suitable choice of t. As s in- creases, the point P describes the curve in a definite sense. Let this be chosen as the positive sense of the tangent line at P. Then t shall be the angle between the positive axis of x and this line. If t were taken as the. angle which the op- positely directed tangent makes with the positive axis of x, the — sign must be written before each right-hand side in (6). The formulas (6) suggest that x and y can be taken as func- tions of s : . , . ... X = ^(s), y = i/f(s). * In case the coordinates x and y are expressed as functions of a third variahle t, dx will not in general be equal to Aa;, but wiU differ from it by an infinitesimal of higher order. The triangle PMQ will then be replaced by a similar triangle PMiQi, in which Mi lies on the line PM, its distance from M being an infinitesimal of higher order. 136 CALCULUS This is, of course, always possible, since, when s is given, P, and hence also x and y, are determined. Since (6) ds = ± \'dx' + dy , we have from (5) (7) sinT = +- — -^^ ■ cosr = ± ^ -s/da? + dy^ Vdx^ + dy'' no matter what choices of s and t are made.* Furthermore, dy (8) 3inT = + — '^^ cosT = ± ^ Which sign is to be used in (8) depends on which of the two possible determinations has been chosen for t. Thus t in a . given cas^ might be 30° or 30° + 180° = 210°. If the first choice were made, t = 30°, then sin. t, cos t, and dy/dx = tan t would all be positive quantities, and hence the upper signs must be taken. But if the other choice, t = 210°, is made, then sin T and cos t are negative, and the lower signs hold. Example. Consider the parabola y = x\ Let P be a point of the curve which lies in the first quadrant. Since tanT = ^ = 2a; dx is here positive, t may be taken as an angle of the first quad- rant. In that case, formulas (8) give 2a! 1 smT = — . cost = Vl + 4a;2 VH-4a;2' * The signs in (6) and (7) are not necessarily tlie same ; also in (7) and (8). TRIGONOMETRIC FUNCTIONS 137 If P is a point of the curve which lies ia the second quad- rant, tan T is negative, and t is an angle of the second or fourth quadrant. If we choose to take t as an angle of the second quadrant, formulas (8) become ■ 2x 1 sinT = . eosT = - Vl + 4aj2 Vl+4a;2 We may, however, equally well take t as an angle of the fourth quadrant. Then sinT= — . cost = - Vl+4a;' Vl + 4a!2 In each ease, one of the numbers, sin t and cos t, is positive, the other, negative. Polar Coordinates. Similar considerations in the case of the curve „.„. lead to the following formulas ; cf . Fig. 46 : PP'^ = P'3P + MP^. limf^Y=limr^%limf^' Hence Now, the chord PP and the arc PP' = As are equivalent infinitesimals. Moreover, PM and Ar are equivalent ; and MP and r^B are equivalent. Hence « Dropping the subscript and writing the derivatives in terms of differentials we have, then : or (10) ds2 = dr-2 + r^dB\ 138 CALCULUS Furthermore, (11) .i«t=f, c<»*.|, the tangent PT being drawn in the direction of the increasiag s, and ^ being taken as the angle from the radius vector pro- duced to the positive tangent. 10. Rates and Velocities. The principles of velocities and rates were treated in Chapter III, § 8. We are now in a position to deal with a wider range of problems. ' We note the following formulas. Let a point P describe the curve „, ^/„v Let s denote the length of the arc, measured from an arbitrary point in an arbitrary sense, and let t be the angle from the positive direction of the axis of x to the tangent at P drawn m the direction of the increasing arc. Then the components of the velocity (v = ds/dt) of P along the axes are, respectively: -H, . da; dy . (1 ) — = v cos T, -s-= V sm T. ^ ^ dt ' dt Let a point P describe the curve (2) r = ¥{&). Let s denote the length of the arc, measured from an arbitrary point ia an arbitrary sense ; and let \ji be the angle from the radius vector, produced beyond P, to the tangent at P drawn iu the direction of the increasing arc. Then the components of the velocity {v = ds/dt) of P along the radius vector pro- duced and perpendicular to the same (the sense of the increas- ing 6 biing taken as positive for the latter) are respectively : ^ ^ dt ^ dt ^ Example 1. A railroad train is running at the rate of 30 miles an hour along a curve in the form of a parabola : y2 = 1000 a;, TRIGONOMETRIC FUNCTIONS 139 the axis of the parabola being east and west, and the foot being taken as the unit of length. The sun is just lising in the east. Find how fast the shadow of the locomotive is moving along the wall of the station, which is north and south, when the distance of the shadow from the axis of the parabola is 300 ft. The first thing to do is to draw a suitable figure, introduce suitable variables, and set down all the data not already put into evidence by the figure. Thus in the present case we have, in addition to the ac- companying figure, the further data : (a) the velocity of the train ;' this must be expressed in feet per second, since we wish to retain the foot as the unit of length for the equa^ tion of the curve. Now, 30 miles an hour is equivalent to 44 feet a second. On the other hand, another expression for the velocity is ds/dt. Hence we have, on equating these two values, ^=44 dt' (b) We must set down explicitly at this "point the equation of the curve, „ . ^^^ To sum up, then, we first draw the figure and then write down' the supplementary data : Given a) -- = 44, dt and' b) y^ = 1000x. The second thing to do is to make clear what the problem is. In the present case it can be epitomized as follows : Tofind (^] ■ We are noj7 ready to consider what methods are at our dis- posal for solving the problem. We observe that ds occurs in 140 CALCULUS the data. Obviously, then, we must make use of the one gen- eral theorem we know which gives an expression for ds when the equation of the curve comes to us in Cartesian coordinates, — namely, the theorem : ds' = da? + dy\ Since dx occurs neither in the data nor in th§ conclusion, we wish to eliminate it. This can be done by means of the equation of the path 6). DifiEerentiating 6) we have : Hence dy = 1000 da;. dx _ydy , 500 Consequently ds^ = "Jf + ( 500" and '=<-i^+^^y- 5002 The next step is obvious ; divide through by dt : ds^ I y I idy dt ^ 5002 "^ at In this last equation, replace ds/dt by its known value from a), and we now have an equation for determining dy/dt : dy 44 \ 5002 ^ Finally, bring into action the particular value of y with which alone the proposed equation is concerned, namely, y-'"'- /^N _ 44 ^ 44 _3,,3^ \dtjy,3a> V.62 + 1 VT36 or, the rate at which the shadow is moving along the wall of the station is 37.73 ft. a second. TRIGONOMETRIC FUNCTIONS 141 Angular Velocity. By the angular velocity, u>, with which a ' line is turning in a given plane is meant the rate at which the angle, ^, made by the rotating line with a fixed line, is in- creasing : do dt Example 2. A point is describing the eardioid r = a (1 — cos 6) at the rate of c ft. a second. Find the rate at which the radius vector drawn to the point is turning when 6 = ir/2. The formulation of this problem is as follows : r — a(l — cos ff).* Given a) and V) To find Since, from § 9 (10), and from 6), ' it follows that ' ds2 = a2 sin2 ed&^ + a\l — cos fff d&^ — a'^de^[sm^6 + 1 — 2 cos ^ + cos^S] =2 d'dff^ • (1 - cos-(9) = 4 a'' sin^ ^ dff^. Hence, s being measured from the cusp, ds = 2asin-dfl, 2 ' J ds rt ■ BdB and — = 2 a sm dt 2 d« * The student should make a free-hand drawing of the curve. 142 CALCULUS Consequently, by the aid of a) ae c ^' 2asinf 2 and thus, finally fde\ _ c »V2 EXERCISES 1. A point describes a circle of radius 200 ft. at the rate of 20 ft. a second. How fast is its projection on a fixed diameter travelling when the distance of the point from the diameter is 100 ft. ? Ans. 10 ft. a second. 2. A flywheel 16 ft. in diameter is making 3 revolutions a second. The sun casts horizontal rays which lie ia or are parallel to the plane of the flywheel. A small protuberance on the rim of the wheel throws a shadow on a vertical wall. How fast is the shadow moving when it is 4 ft. above the level of the axle ? 3. A revolving light sends out a bundle of rays that are approximately parallel, its distance from the shore, which is a straight beach, being half a mile, and it makes one revolu- tion in a minute. Find how fast the light is travelling along the beach when at the distance of a quarter of a mile from the nearest point of the, beach. 4. A point moves along the curve r = 1/6 at the rate of 6 ft. a second. How fast is the radius vector turning when 6 = 2w? 5. In the example of the ladder. Chap. Ill, § 8, Ex. 5, find how fast the ladder is rotating at the instant in question. 6. At what rate is the direction of the second ship from the first changing at the instant in question, in Ex. 2 of Chap. Ill, §8? TRIGONOMETRIC FUNCTIONS 143 7. How fast is the direction of the man from the lamp- post changing in Ex. 12 of Chap. Ill, § 8 ? 8. The sun is just setting aa a baseball is thrown vertically upward so that its shadow mounts to the highest point of the dome of an observatory. The dome is 50 ft. in diameter. Find how fast the shadow of the ball is moving along the dome one second after it begins to fall,' and also how fast it is moving just after it begins to fall. 9. Let AB, Fig. 48, represent the rod that connects the piston of a stationary engine with the fly-wheel. If u denotes the velocity of A in its rectilinear path, and V that of B in its circular path, / «/V^~-— i show that M =:(sin 6 + cos 6 tan ^)?;. FiQ. 48 10. rind the velocity of the piston of a locomotive when the speed of the axle of the drivers is given. 11. A drawbridge 30 ft. long is being slowly raised by chains passing over a wind- lass and being drawn in at the rate of 8 ft. a minute. A distant electric light sends out horizontal rays and the bridge thus casts a shadow on a vertical wall, consisting of the other half of the bridge, which has been already raised. Find how fast the shadow is creeping up the wall when half the chain has been drawn in. i 12. A man walks across the floor of a semicircular rotunda 100 ft. in diameter, his speed being 4 ft. a second, and his path the radius perpendicular to the diameter joining the extremities of the semicircle. There is a light at one of the latter points. Find how fast the man's shadow is moving along the wall of the rotunda when he is halfway across. Fig. 49 144 CALCULUS 13. A man in a train that is running at full speed looks out of the window in a direction perpendicular to the track. If he fixes his attention successively for short intervals of time on objects at different distances from the train, show that the rate at which he has to turn his eyes to follow a given ohject is inversely proportional to its distance from him. 14. Water is flowing out of a vessel of the form of an in- verted cone, whose semi-vertical angle is 30°, at the rate of a quart in 2 minutes, the opening being at the vertex. How fast is the level of the water falling when there are 4 qt. of water still in ? 15. Suppose that the locomotive of the first of the Examples worked in the text is approaching the station at night at the rate of 20 miles an hour, its headlight sending out a bundle of parallel rays. How fast will the spot of light be moving along the wall of the station when the distance of the head- light from the vertex A of the parabola, measured in a straight line, is 500 ft. ? Assume that the wall is perpendicular to the axis of the parabola and distant 75 ft. from the vertex. 16. In the preceding question, how fast will the bundle of rays be rotating ? 17. A point describes a circle with constant velocity. Show that the velocity with which its projection moves along a given diameter is proportional to the distance of the point from this diameter. 18. A point P describes the arc of the ellipse which lies in the first quadrant, at the rate of 12 ft. a second. The tangent at F cuts off a right triangle from the first quad- rant. How fast is the area of this triangle changing when P passes through the extremity of the latus rectum ? Is the area increasing or decreasing ? TRIGONOMETRIC FUNCTIONS 145 19. A point P describes the oardioid r = 5 (1 — cos ff) at the rate of 12 cm. a second. The tangent at P cuts the axis of the curve in Q. How fast is Q moving when 6 = 7r/2 ? 20. The sun is just setting in the west as a horse is running around an elliptical track at the rate of m miles an hour. The axis of the ellipse lies in the meridian. Find the rate at which the horse's shadow moves on a fence beyond the track and parallel to the axis. CHAPTER VI LOGARITHMS AND EXPONENTIALS 1. Logarithms. The logarithms with which the student is familiar are those which are ordinarily used for computation. The base is 10, and the definition of logioic is as foUows : y = logio X if 10" = X. These are called denary, or Brigg^s, or common logarithms. More generally, any positive number, a, except unity, caai be taken as the base, the definition of log, x then being : (1) 3/ = log, a; if a«=x. y = log I Fig. 50 The accompanying figure represents in character the graph of the function log, x for any a > 1. It is drawn to scale for 146 LOGARITHMS AND EXPONENTIALS 147 a = 2.71828. The reason for this choice of a will appear shortly. From the definition it follows at once that (2) log.l = 0, log„a = L Only positive numbers have logarithms. For, a" is always positive. Hence, if x be given a negative value (or the value 0), the second equation under (1) above cannot be satisfied by any value of y. The two leading properties of logarithms are expressed by the equations : * (I) logP+logQ = log(PQ) (II) logP" = nlogP. Here, P and Q are any two positive numbers whatever, and n is any number, positive, negative, or zero. The base, a, is arbitrary. Thus log 10 = log 2 + log 5 and log VT = log 7^ = J log 7. From equation (I) it follows that (3) log^ = -logQ and (4) log 1= log P- log Q. For, if we set P = 1/Q in (I), we have But, by (2), log 1 = log - + log Q. log 1 = 0. * The student should recall the proofs of these theorems, which he learned in the earlier study of logarithms, and make sure that he can reproduce them. Proofs of the theorems are given in the author's DWerential and Integral Calculus, p. 76. 148 CALCULUS Hence i 1 i /-» j log- = -logQ, q.e.d. Again, write (1) in the form logiPQ') = logP+logQ', and now set Q' = 1/Q. Then P i log- = logP+log-. But logi = -logQ. Hence p log-=logP-logQ, q.e.d. For example, log (a + b)- log a = log fl + ^\ as we see by setting, in equation (4), P=a + b, Q = a. As a further example of the application of equation (II) we may cite the following : ^°g ^t + ^) = log Ka + ¥}- ft Eor, if P=a + 6 and »=-, the left-hand side of this equa^ h tion has the value n log P. A Further Property of Logarithms. When it is desired to express a logarithm given to a certain base, a, in terms of logarithms .taken to a second base, b, the following relation is needed : (III) log.a; = j5Si^. The proof of (III) is as follows. Let y = log» X, a' = X. LOGARITHMS AND EXPONENTIALS 149 Take the logarithm of each side of this equation to the base b : (5) logj a» = logt X. But the left-hand side can be transformed by (11)^ if in (II) we take 6 as the base, thus having logjP»=nlogiP. Here, let P=a, n = y. Then logj a<'=y log^ a, and (5) now becomes : y logj a = logi X. y = ^^, or log„., = |2i^, q.e.d. logs a logs* Example. Let 6 = 10 and let a = 2.718. To iind log„ 2. From (III), log 2 = _i2gio2_ _ .3010 ^ gg32 ^° logio 2.718 -4343 ' Two Identities. Just 'as, for example, -\/q^ = x and (Va:)' = «, no matter what value x may have, so we can state two identi- ties for logarithms and exponentials. In the second equation (1), replace y by its value from the first equation. Thus the equation (6) rf»So"- = a! is seen to hold for all positive values of x. Secondly, replace x in the first equation (1) by its value from the second equation : y = log„ a". We can equally well write x instead of y, understanding now by x any number whatever, and we have, then, the 150 CALCULUS identity (J) log. a' = X. This equation holds for all values of x, positive, negative, or zero. EXERCISES 1. Show that , -.„. -._„ logio .8950 = — .0482. 2. rind logio .09420. -4ns. -1.0259. 3. Compute 2.718-6M2. Ans. 1.758. 4. Compute 2.718--«'^°. Ans. 0.4186. 5. Compute w''. 6. Compute V2*'^_ 7. Show that log tan 6 = log sin 6 — log cos ^, <6 <-■ 8. Show that log sin e + log cos fl = log 55:^, o<e<l. 9. Show that j^gl-|ose_21ogsin|, 0<fl<2,r. 10. If (a;, y) are the Cartesian coordinates of a point distinct from the origin, and (r, ff) the polar coordinates of the same point, show that , , , , „ « ^ ' log r = 1^ log (as' + 2^'). 11. Prove that log (a^ - 62) = log (a + 6) + log (a - 6), provided a + 6 and a — b are both positive quantities. 12. Simplify the expression log (1 + ;»•)- log (l+a!2). 13. Show that V(e^ — e-')2 + 4 = e' + e"*, where e has the value 2.7182. LOGARITHMS AND EXPONENTIALS 151 14. Simplify the expression 15. Show that ilog(l + = log(H-0'. 2. Differentiation of Logarithms. In order to differentiate the function , it is necessary to go back to the definition of a derivative, Chap. II, § 1, and carry through the process step by step. Give to X an arbitrary positive value, Xq, and compute the corresponding value, 2/0, of the function : (1) VO = loga Xg. Next, give to x an increment Ax (subject merely to the restric- tion that Xq + Ax is positive and Ax # 0) and compute the new value, 2/0 + Ay, of the function : (2) yo + Ay= log„ (xa + Ax). From (1) and (2) it follows that Ay ^ log, (xq + Ax) — log„ x„ Ax Ax It is at this point that the specific properties of the loga- rithmic function come into play for the purpose of transform- ing the last expression. By § 1, (4), log„ {xo + Ax) - log„ xo = log„ fl + ^\ and hence We next replace the variable Ax by a new variable t as follows : 152 CALCULUS t=— or Aa; = x^. Xf, Thus (3) takes on the form From (II), § 1, the bracket is seen to have the value log. (1 + OS and hence ^ = ilog„(l + 0'. (4) Ax Xg As Aa; approaches as its limit, t also approaches 0, and so (6) limf2^ = llimlog,(l + 0^- i_ Now, the variable (1 + ' approaches a limit when * ap- proaches 0, and this limit is the number which is represented in mathematics by the letter e; cf . § 3. Its value to five places of decimals is e = 2.71828 .•• ; cf . § 3. Moreover, log a; is a continuous function of x, as is shown in a detailed study of this function.* Hence lim log„ (1 + 0'"= log„ { lim (1 + *)4 = log« «■ On substituting this value in the right-hand side of (5) we have: Xq * Such a treatment is too advanced to be pursued with profit at this stage. Cf. the author's Differential and Integral Calculus, Appendix, p. 417. LOGARITHMS AND EXPONENTIALS 153 or, on dropping the subscript : (6) D,log„a; = ?:5g^. X Thus if the usual base, a = 10, be taken, the formula becomes : f7\ n 1 .4343... (7) Z)Jogioa!= X Discussion of the Result. We have met a similar situation before, in the differentiation of the sine. There, if angles be measured in degrees, the fundamental formula reads : J5, sin X = -^ cos x. 180 In order to get rid of this inconvenient mxdtiplier, we changed the unit of angle from the degree to the radian, and then the formula became : Dj sin X = cos x. In the present case, it is possible to do a similar thiag. The base, a, is wholly in our control, to choose as we like. Now, for any base, the logarithm of the base is unity, § 1, (2) : log„ a = 1. If, then, we choose as our base the number e : a = e = 2.71828 - the multiplier becomes (8) log„ e = log. e = 1. For this reason, e is taken as the base of the logarithms used in the Calculus.* These are called natural logarithms. They are also called hyperbolic, or Naperian logarithms, — the latter name after Napier, the inventor of logarithms. But * The notation e for this number is due to Euler, 1728. 154 CALCULUS Napier * was the very man who introduced denary logarithms into mathematics, and so the use of his name in connection with natural logarithms is misleading. Since natural "logarithms are always meant in the formulas of the calculus, unless the contrary is explicitly stated, it is customary to drop the index e from the notation log, a; and to write (9) y = log X, if e" = x. The identities (6) and (7) of § 1 now take on the form : (10) 6'°^" = X, (11) log e' = X. The formula of differentiation becomes : (12) DJogx = -- X In differential form it reads : (13) 4-logx=^, dx X (14) dlo8x = ^. X Example. Differentiate the function u = log sia X. Let y = sin X. Then u = logy, t Z«. = rf;ioB-i/=^. d dy = cos xdx, y and , cosxdx . , du = — : = cot X ax. siniB * Napier was a Scotchman, and Ms discovery was published in 1614. LOGARITHMS AND EXPONENTIALS 155 Hence d log sin x = cot x dx, or — log sin x = cot x. dx EXERCISES Differentiate the following functions. 1. M = log cos X. -r- = — tan x. dx 2. M = log tan x. — = cot x + tan x. die du -2 3. M = log cot «. , — . r> aa; sinzoj 4. M = log sec K. 5. M = log CSC a;. „ , 0! du 1 , 1 6. u = log:j — =- + 7. M = log 1 — a; da!a!l — K a + a; du _ 2a a — a; da; a^ _ a;2 (Zm a; 8. M = log Va^ + xK _ 1 /< \ dw . a; 9. M = log (1 — cos a;). — = cot - da; 2 10. u = log (1 + cos a;). T'~~ ^^^ n dx a? + x^ du .x — = cot — da; 2 du , — = — tan da; 2 3. The Limit lim (1 + t) '- Since this limit is fundamental in the differentiation of the logarithm, a detailed discussion of it is essential to completeness. Let us set (1) s=(l + 0^ and compute the value of s for values of t near 0. Suppose t = .1. Then s= (1.1)10, 156 CALCULUS and this number is found by the usual processes with loga- rithms to be 2.69. Further pairs of corresponding values {t, s) are found in a similar manner. In particular, the student can verify the correctness of the following table of values : * t -0.1 -.01 -.001 . . + .001 + .01 + 0.1 s 2.87 2.73 2.72 . . 2.72 2.70 2.59 The foregoing table indicates strongly that, when t ap- proaches the limit from either side, the variable s is approaching a limit whose value, to three significant figures, is 2.72. This is ia fact the case.f The exact value of the limit is denoted by the letter e : I (2) lim (1 + «)' = e = 2.71828 -. 4. The Compound Interest Law. The limit (2) of § 3 pre- sents itself in a variety of problems, typical for which is that of finding how much interest a given sum of money would bear if the interest were compounded continuously, so that there is no loss whatever. For example, $ 1000, put at in- terest at 6%, amounts in a year to $1060, if the interest is not compounded at all. If it is compounded every sis months, we have , f^/., $1000(1 -I- 4^ j as the amount at the end of the first six months, and this must be multiplied by [l4-^| to yield the amount at the end of the second sis months, the final amount thus being $iooo/'i + --^^ * To compute the middle entries in this table a six-place table of logarithms is needed. t For a rigorous proof cf . the author's Differential and Integral CaJr cuius, p. 79. LOGARITHMS AND EXPONENTIALS 157 It is readily seen that if the interest is compounded n times in a year, the principal and interest at the end of the year will amount to , „„. „ 1000(^1 + :^) dollars, and we wish to find the limit of this expression when «. = 00. To' do so, write it in the form : 1000 n and set t=' — . The bracket thus becomes n and its limit is e. Hence the desired result is 1000e»« = 1061.84.* EXERCISE If $ 1000 is put at interest at 4 % , compare the amounts of principal and interest at the end of 10 years, (a) when the interest is compounded semiannually, and (6) when it is com- pounded continuously. Ans. A diEEerence of $ 6.88. 5. Differentiation of e'. Before beginning this paragraph the student wUl turn to Chap. VIII and study carefully § 1. Since (1) y = ^ and X = log y are equivalent equations, the former function can be differen- tiated by taking the differential of each side of the latter equation: ^ ax = diogy = -^- y * The actual computation here is expeditiously done by means of series ; see the chapter on Taylor's Theorem. 158 Hence or (2) (3) The function CALCULUS dy _ dx y, d^ dx «*, de = e'cte. y = e' . .1 X Fig. 51 ' = a' (4) could be differentiated in a similar manner. It is, however, simpler to take the logarithm of each side of (4) and then dif- ferentiate the new equation : logy = log a* = a; log a, d log w = -=^ = da; log o. y (5) da' = al'logadx. Differentiation of a!". It is now possible to complete the dif- ferentiation of this function for the case that n is irrational. LOGARITHMS AND EXPONENl Since by § 2, (10), X = e'-s', it follows that x' = e"^"^', and hence = e''^''^'d(nlogx) -eni^e^ndx X Thus finally, X (6) dx'" = «a!"~idx, 159 no matter what value n may have, provided merely that n is a constant. Differentiation of /(a!)'"'''. Let it be required, for example, to differentiate the function 2/ = af. Here, both base and exponent are variable. Begin by tak- ing the logarithm of each side of the equation : log y = log X' = X log X. Hence ,, ,, , a log y = d{fe log x), °^ ^ = {l + \ogx)dx, y and so, finally, das' = a?(l + log x)dx. dy = y{l + log X) dx or The general case, „ _.y(^a;')*' can be treated in a similar manner. 160 CALCULUS 6. Graph of the Function «". Por positive values of n the curves y = x' Fis. 52 lie as indicated in the figure. When n = 1, we have the ray from the origin, which bisects the angle between the positive axes of X and y. LOGARITHMS AND EXPONENTIALS 161 When n > 1, the curve is always concave upward ; when w < 1, it is concave downward. All the curves start at the origia and pass through the point (1, 1). Por values of a; > 1, the larger n, the higher the curve lies. For values of a; < 1, the reverse is the ease. Let X have any fixed value greater than unity : a; = a;' > 1. Consider the ordiaate ,„ y = x'". As n increases, x'" increases continuously. This property is the basis of the property of logarithms included m the word continuous. For proofs of the foregoing statements of. the author's Differential and Integral Calculus, p. 27 and Appendix, p. 417. 7. The Fornmlas of Differentiation to Date. The student will now bring his card of formulas up to date by supple- menting it so that it will read as follows : Genebal Formulas of Diffbrbsttiation I. dcu = c du. IL d(u + v) = du + dv. III. d (uv) =udv + v du. TV ,fu\_ vdu — udv \vj ifl Special Formulas of Differentiation 1) dc = 0. 2) da;'' = wa;"~ida;. 3) d sin X = cos. x dx. 4) d cos a; = — sin x dx. 5) dtan x = sec^ x dx. 162 CALCULUS 6) dcota! = — csc^a; da;. 7) dloga; = ^. X 8) d^ = ^dx. 9) da" = a" log a dx. To obtain facility in the use of the new results it is desirable that the student work a good number of simple exercises. Example 1. To differentiate the function u = e"'. Let y = ax. Then u = e", du = de" = e''dy = e"(adx). Hence -"'■ = ae' ' de" = ae°*da; or — e' dx Example 2. show that * If u = Aco8 (nt + y), To do this, compute first — . The computation is readily az effected by taking the differential of each side of the given equation : du = Ad(X)s(nt+y) = A£—sia{nt + y)d(nt+ y)] = — An sin (nt + y) dt, * Such an equation as the following is called a differential egvation, and any function which, when substituted for u, satisfies the equation is called a solution. LOGARITHMS AND EXPONENTIALS 163 — = — An sin {nt + y). Next, compute — . Since dH_d^du\ dP~di\dt) ~~dF dM_ dt' / dt we take the differential of each, side of the equation for df — - J = — An d sin (nt + y) — — An [cos {nt + y)d{nt + y)] = — An^ cos (nt+y) dt. Hence, on dividing through by dt, we have : ^ = -An^ COS (nt + y). If now we multiply the given value of u by n^ and add the dhi product to the value just obtained for — , the result is iden- tically 0, i.e. for all values of t : — + n^M = 0, q. e. d: dt^ ' ^ EXERCISES Differentiate the following functions. 1. u = e-''. ~ = -2xe-". dx dit 2. w = e='"^ — =e™»^cosa;. da; 3. M = (e» + e--)2. ^ = 2(e2'-e-2'). dx 164 CALCULUS 4. M = 10'. — = (2.30259 OlO*. da; 5. w = a!i<'10'. ^=»9 10-(10 + 2.30259a!). dx 6. u = log (sec X + tan x). — = see x. dx 7. M = a!2 1oga;. — = a;(l + 2 loga;). dx 8. M = a' log (a — a;). 9. m = e"'log(2a; + 3). 10. u = e""' COS (nt — y). 11. u = e~«*(^ cos ni-f-B sin n<). a; log a; , / , i\ dw loga; 12. M= — i^:^ log(a; + l). — = . f^, • a; + l da; (a; + 1)2 dw 1 13. u = log(a;+Va;2_a2). ^ = 14. u = log (as + Va^ + a 15. u = log (e* + e^). 17. M = log tan- • 18. M = logtan^| + j\ da; Va;" — a? du 1 19. „ = cotg-| 20. u 21. M = log VI + sin 0. 23. u = log Vl — cos X. 25. M=(10i+')=. 27. «=r^Y- da; Va2 + a;'' 16. M = sin a; + cos x e' du dx' = CSC X. du do'' = sec 6. du 1 dx 1 — sin a; du 1 dx 1 + sina; 22 . M = = log«^^. X 24 . U = =v^. 26 . Mi = </^. 28, , U = = VIO'. LOGARITHMS AND EXPONENTIALS 165 29. u = af"". ^ = a!'"'»'-i(sina; + a!cosa!loga;). 30. M = (siiia;)'=°". ^= (suia!)™»--i(cos2a!- sitfajlogsina;). 31. u = a;"'. 32. u = (cos a;)*'""^. 33. u — (tan x)". X 34. •M = (loga;2)^ 35. m = (1 + a)«. 36. u = {x^f. 37. If M = J. cos ni + JB sin nJ, show that 38. If M = Oe-" cos (Vw^ — K^f + y), show that h 2 K h w% = 0. d«2 df CHAPTER VII APPLICATIONS 1. The Problem of ITninerical Compntation. It often happens in practice that we wish to solve a numerical equa- tion in one unknown quantity, or a pair of simultaneous equations in two unknowns, to which the standard methods with which we are familiar do not apply ; for example, cos x = X, f 2 cot fl + 2 = cot <^, [ 2 cos + cos <f> = 2. Such equations usually come to us from physical problems, and the solution is required only to a limited degree of accuracy, — say, to two, three, or possibly four significant fig- ures. Any method, therefore, which yields an approximate solution correct to the prescribed degree of accuracy furnishes a solution of the problem. In particular, the problem of the determination of the error in the result due to errors in the observations comes under this head. 2. Solution of Equations. Known Graphs. Example 1. Let it be required to solve the equation 1) cos x = x. We can evidently replace this problem by the following: To find the abscissa of the point of intersection of the curves 2) y = cos x, y = x. 166 APPLICATIONS 167 The first of these curves we have plotted accurately to scale. The second is the right line through the origin, which bisects the angle between the positive coordinate axes. It is, there- fore, sufficient to lay down a ruler on the graph of the former curve, so that its edge lies along the right line in question, and observe where this line cuts the curve. The result lies between _ , x= .1 and X = .8, and may fairly be taken as a; = .76. It is understood, as usual iu approximate values, that the last figure tabulated does not claim complete accuracy ; but we are entitled to a some- what better result than would be given by the first figure alone. 'Example 2. To solve the equation 3) a!3-f-2a;-2 = 0. Suppose we have plotted the curve 4) y = 0? accurately from a table of cubes. Then the problem can con- veniently be formulated as follows : To find the abscissa of the point of intersection of the curves 5) y — s? and y — 2 — 2x. The details are left to the student. Example 3. To find the positive root of the equation 6) e-i''4- 2.92 a; = 2.14. Here, we can connect up with the graph of the function e* by making a simple transformation. Let 7) x'=: — ^x; x = — 2x'. The equation then becomes 8) es'' - 5.84a;' = 2.14, 168 CALCULUS and we seek to determine the abscissa of that point of inter- section of the curves (for simplicity, we drop the accent) 9) ^ = 6^ and y = 5.84a; + 2.14 which lies to the left of the origin. The second place of decimals in the coefllcients is not to be taken too seriously ; we make as accurate a drawing as the graph and a well-sharpened pencil permit. Having thus determined the negative a' from the graphs of 9), we find the desired positive x by substituting this value in equations 7). The execution of the details is left to the student. Example 4. Solve the equation e* = tan x, < a; < ~ / If one of the curves 2/ = e* or y=. tan x were plotted on transparent paper, or celluloid, it could be laid down on the other with the axes coinciding and the inter- section read off. The same result can be attained by holding the actual grapTis up in front of a bright light. In cases as simple as this, however, free-hand graphs will often yield a good first approximation, and further approxima- tions can be secured by the numerical methods of the later paragraphs. EXERCISES* 1. Solve the equation cos X = 2x. 2. Find the root of the equation 3 sin x = 2x which lies between and tt. * In solving these exercises only so great accuracy is expected as can be attained from well-drawn graphs of the standard cviTves. It will be shown in later paragraphs how the solutions can be improved analytically and carried to any desired degree of accuracy. APPLICATIONS 169 3. Solve: X + tan a; = 1, o<.<|. 4. Solve: 3 cos a; — 6a! = 6, -|<.<o. 5. Find the root of the equation log x^! + 2 = a; which lies between and 1. 6. Solve : sin 2a! = x. 7. Find all the roots of the equation 12a!' + 4a! + 3 = 0. 8. The same for „ , „ ^ r. K)'ie — t>x — 1 = 0. 9. The same for ar* — a; — 1 = 0. Solve the following equations : 10. cos' e + .47 cos - 1.23 = 0, < 6 < 90°. 11. sin X = Vl — x^. 12. a;2 + cos^ a; = 4. 13. Show that the equation tan a; = a; has an infinite number of roots. These can be written in the form x„ = Wff + e„, where e„ is numerically small when n is numerically large. 14. Find the largest value of P for which the equation cos X + Px = 1 admits a solution in the interval < a; < ir. 15. Find the point of the parabola 2y = x^ which is nearest to the point (2, 0). 170 CALCULUS 16. rind the radius of the circle whose center is at (0, 2) and which is tangent to the parabola y^ = x. 3. Interpolation. Consider the equation 1) f{x)=0. Suppose a root has been located with some degree of accuracy. More precisely, suppose that /(rBi) and f{x^ are of opposite signs. If the function f(x) is continuous in the interval Xi'^x^x^ and if its derivative is always posi- tive (or always negative) in this interval, then the function is always increasing (or always decreasing) and so must have just one root between asj and x^. The root can be found approximately as follows. Consider the graph of the function 2) y=m. Let 2/i =/(»!), Vi =f{x2), and draw the chord through the points (a^, y{) and (ajj, y^. The point in which this chord cuts the axis of x will obviously yield a further approximation to the root sought. Denote this last value by X. The equation of the chord is 3) Jg - a^i ^ y - 2/1 ■ s^ — ^i 2/2 — 2/1 Fig. 53 On setting 2/ = and solving for x, we have, as the value of X, the following : 4) X=xi-?i^^2,j, 2/2-2/1 or 5) X=x,--^^^^^^f(x,). APPLICATIONS 171 We have explained the method in detail and developed, in equations 4) and 5), the analytic formula for the determina- tion of the new approximation, X. In practice, however, it is usually simpler to draw the straight lines of Fig. 63 accurately on a generous scale and read off from the figure the value of X. Example. Consider equation 3) of § 2, Ex. 2 : 6) a' + 2a; - 2 = 0. The curve in question is here 7) y = ay' + 2x — 2, and the graphical solution of § 2 shows that the root is about a; = .7 or .8. Let a; = aji = .7 ; then j/i is found to have the value 2/1 = -.267. Next, let x = X2 = .S; then y^ = .112. We have, then, to lay a secant through the points i^i, Vi) = (-7, - .257) and {x„ y,) = (.8, .112). Its equation is given by 3) * : a; -.7 ^ y + .257 .8 -.7 .112 + .257' On setting y = in this equation and solving for x, we get, ''.' X=.7+^^ = .7693. .369 In order to see about how close this approximation is, com- pute the corresponding value of y : 2/|z=.7693 = — -0063. We get, then, about two places of decimals, x = .77. * It is desirable that the student should make this determination graphically, as indicated above in the text. He should take 10 cm. to represent the interval of length .1, from Xi = .7 to aj = .8. 172 CALCULUS It is possible to apply the method again, taking now (ao, 3/i)=(.7693, - .0063) and (a^, ^2) as before. We leave this as an exercise to the student. He should make both the graphical determination with an enlarged scale and the analytic determination of formula 4). The Method; Not, the Formula. The student may be tempted to use the formula 4) or 5), rather than to go back to the method by which it was derived. This would be un- fortunate, for the formula is not easily remembered, whereas the method, once appreciated, can never be forgotten. If the student finds himself in a lumber camp with nothing but the ordinary tables at hand, he may solve his equation if he has once laid hold of the method. It is true that the best way is for him to treat first the literal case and deduce the formula. But this he may not be able to do if he has relied on the formula in the book. EXERCISES Apply the method to a good number of the problems at the end of § 2. 4. Newton's Method. Suppose again that it is a question of solving the equation 1) f{x) = Q, and suppose we have already succeeded in finding a fairly good approximation, x = Xi. Consider the graph of the function 2) y =/{<>=)■ Compute Hi =f(x,). To improve the approximation, draw the tangent at the point (xi, y^. Its equation is : APPLICATIONS 173 Evidently, this luie will cut the axis of a; at a poiat very near the point in which the curve 2) cuts this axis. If, then, we set y = in 3) and solve for x, we shall ob- y tain a second approximation to the root of 1) which we seek. The value of this root will be 4) X = xi- Fig. 54 Example 1. Let us apply the method to the Example studied in § 3. In order, however, to have simpler numbers to work with, take Xi = .77 and compute the corresponding yi ; it is found to be : j/j : .0035. («i,2/i) = (-77, -.0036). We must next compute dy/dx from the equation y = a^ + 2x—2; dx (dy\ =3.779. On substituting these values in 3), we have : 2/ + .0035 =3.779 (a; -.77). Now set 2/ = and solve. The result is that given by 4) : .0035 a; = .77 + 3.779 " .7709. We have tabidated four figures in the result because this is about the degree of accuracy that seems likely. To test this point, compute y for the value of x which has been found : 3/U=.7709 = --000L Since the slope of the graph is greater than unity, the error in a; is less than one unit in the fourth place. It is easy to verify the result by computing y for the next larger four-place value of a.: y |^..„,„ = + .0003. 174 CALCULUS Thus we have a complete proof that the root lies between . 7709- and .7710, and we see that it lies about one quarter of the way from the first to the second value. Example 2. It is shown that the equation of the curve in which a chain hangs, — the Catenary, — is 5) 2,=|^e^ + e-^ where a is a constant. The length of the arc, measujed from the vertex, is 6) Let it be required to compute the dip in a chain 32 feet long, its ends being supported at the same level, 30 feet apart. We can determine the dip from 5) if we know a, and we can get the value of a from 6) by setting s = 16, x = 15: 16 Leta; = — . Then / 15 15, f(x) = ei'-e-'-^x = 0, and we wish to know where the curve 7) y=f{x) = e-e-'-^x crosses the axis of x. This curve starts from the origin and, since ax is negative for small values of x, the curve enters the fourth quadrant. Moreover, ^=6^-6-- >0, x>0, APPLICATIONS 175 and hence the graph is always concave upward. Einally, f(l) = e-e-^-2^ = .217>0, and so the equation has one and only one positive, root, and this root lies between and 1. It will probably be better to locate the root with somewhat greater accuracy before beginning to apply the above method. Let us compute, therefore, /(^). By the aid of Peirce's Tables we find : /(.5) = 1.6487 - .6066 - 1.0667 = - .0245 < 0. Comparing these two values of the function : /(.5) = -.02, /(I) = .22, and remembering that the curve is concave upward, so that the root is somewhat larger than the value obtained by direct interpolation (this value corresponding to the intersection of the chord with the axis of a;) we are led to choose as our first approximation Kj = .6 : /(.6) = 1.8221 - .5488 - 1.2800 = -.0067, /'(.6) = 1.8221 + .5488 - 2.1333 = .2376. Hence the value of the next approximation is X = .6 - z:j99K = .6 + .0282 = .628. .2376 To get the next approximation we compute /(.628) = 1.8739 - .5337 - 1.3397 = .0005. Hence the value of the root to three significant figures is .628 with a possible error of a unit or two in the last place, and the value of a we set out to compute is, therefore, 15/.628 = 23.9. Remark. Newton's method, like the other methods of this chapter, has the advantage that an error in computing the new approximation wUl not be propagated in later computations. Such an error wUl in general hinder us, because we are not 176 CALCULUS likely to get so good an approximation. But the one test for the accuracy of the approximation is the accurate computa- tion of the corresponding y, and if this is done right, we see precisely how close we are to the desired root. The function /(») is usually simple, and it is easy to see whether the curve is concave upward or concave downward near the point where it crosses the axis. We thus have a means of improving the approximation at the same time that we simplify the new value of x. For, if the curve lies to the right of its chord, the approximation by interpolation will be too small ; and if the curve lies to the right of its tangent be- tween the point of tangency and the axis of x, the approxima- tion given by Newton's method will also be too small. Comparison of the Two Methods. When looked at from their geometric side the two methods appear much alike, the first seeming somewhat simpler, since it does not involve the use of derivatives. Why bother, then, with Newton's method? It is not a theoretical question, but purely one of convenience in carrying out the numerical work. It will be found that, as a rule, the first method is preferable in the early stages (usually, merely in the first stage). When, however, a fairly good approximation has been reached, the numerical work ra- volved in Newton's method is generally shorter than that required by interpolation. EXERCISES Apply the method to the Exercises of § 2. When, however, the approximation given by the graphical method of § 1 is crude, the method of interpolation may be used to improve it. 5. Direct Use of the Tables. Example 1. Let us recur to the first example studied, Ex. 1, §2: 1) cos a; = a?. APPLICATIONS 177 The graphical solution gave x = .75. Turn now to a table of natural cosines in radian measure, preferably Peiroe's Tables. As we run down the table, we find the entries : RADIANS .7389 .7418 cos NAT .7392 .7373 Thus X is seen to lie between .7389 and .7418. It is an ex- cellent exercise for the student to work out the interpolation for himself before we take it up at the end of the paragraph. The answer is : a; = .7391. Example 2. Consider the equation 2) tan X = e', the desired root lying between and Tr/2. A free-hand drawing of the graphs of the functions y = tan x, y=e'' shows that x lies between 1 and 1.5. So the next step is taken conveniently by opening Peirce's Tables to the Trigonometric Functions and Huntington's to the Exponentials, and writing down the two pairs of values of the functions which came nearest together : 1.3 1.4 tan X 3.60 5.80 3.67 4.06 Thus the root is seen to lie between 1.3 and 1.4. The general case which the above examples are intended to illustrate is the following : — To solve thf equation f{x) = 4>{x), where f{x) and <^{x) are tabulated functions, or functions readily computed. 178 CALCULUS When the solution has progressed to the point indicated by the examples, the next step can be taken by interpolation, or by Newton's method, as will now be explained. Interpolation. When two values of the independent variar ble near together, Xi and x^, have been found such that /(») is greater than <^(a!) for one of them and less than ^(x) for the other, the best approximation to take next is the one given by the abscissa of the point of intersection of the chords of the graphs of the functions. This value, X, can be found as follows. Suppose that /(oh) < <^(a;i) and /(ajj) > (^(k^). Introduce the following notation : X — Xi^ h. From the figure, the triangles AiCBi and A2CB2 are similar, and AiBi = Ai, A^Bi = Aj. Their altitudes, when C is taken as the vertex, are respectively h and & — h. Hence h ^ B-h Ai Aj On solving this equation for h we find : 3) A = - A, A1 + A2 8. If /(a!i) > ^(xi) and/(a^) < ^(3:2), the result still holds, for Ai and A2 now become negative, but their numerical values APPLICATIONS 179 correspond to the lengths of the sides of the triangles in question. It is easy to express in words the result embodied in 3). Etjle. In order to see what fraction of 8= x^ — Xi must be added to x^ in order to give X, form the differences Then the fraction is the quotient of the first of these differences by their sum. In practice, an accurately drawn figure on a large scale will often afford a quicker and sufficiently accurate solution. Example. Eetiiming to Ex. 1 above, we have : f(x) = cos X, ^(^)= ^ ; 8 = X2-Xi = .0029, «! = .7389, % = .7418. 4>(x{)-f{x0=-.OOO4: ; f{x^)- <t>(,x^)=- .0045. .0004_oo29 = :0116^_0002. .0049 49 Hence the value of the new approximation is X = .7389 + .0002 = .7391. The student will have no difficulty in completing Ex. 2 above in a similar manner. It turns out that the correction is here less than one tenth of 8, and hence it does not influence the second place of decimals : x' = 1.30. Newton's Method. If a higher degree of accuracy is desired, it is well now to apply Newton's method to the function F{x)=f{x)- <l>{x). In the case of Ex. 1 above it is pretty clear that we already have four-plaee accuracy, and the computation of F{x) for the value X = .7391 would only verify the result. This is as far as we can go with four-place tables. If we needed greater 180 CALCULUS accuracy, we should use Newton's method and five or six-place tables. Example 2 has been carried only to two-place accuracy, or three significant figures. We can obtain two further figures with the tables at our disposal, y = F{x)= tan x — e'. 2/1 = F(1.30)= 3.602 - 3.669 = - .067. ^ = sec^a;-e^ ^ dx dx = 13.97 - 3.67 = 10.30 X=1.30-f — = 1.3067. 10.3 To test this result, however, would require five-place tables. EXERCISES Solve the following equations : 1. cot a; = as, < a; < tt. 2. e* -|- loga!= 1. 3. The hyperbolic sine (sh x or sinh a;) and cosine (ch x or cosh x) are defined as follows : sha5 = — - — , cha! = --L — , and are tabulated in Peirce's Tables, pp. 120-123. By means of these, reduce the treatment of Ex. 2, § 4, to the methods of the present paragraph. 6. Successive Approximations. We come now to one of the most important of all the methods of numerical computation. In physics it is known as the method of Trial and Error ; in mathematics it goes under the name of the method of Succes- sive Approximations. The problem is that of solving a pair of simultaneous equations, 1) F{x,y)=0, ^(x,y)=0. APPLICATIONS 181 The cases which arise in practice are characterized in general by t-wo things : First, there is only one solution of the equa- tions -which interests us, and the physical problem enables us to make a fairly good guess at it for the first approximation. Secondly, each of the equations 1) is simple, the curve can readily be plotted in character, and the equation can be solved with ease numerically for the dependent variable when a nu- merical value has been given to the independent variable. But elimination of one of the unknowns, though sometimes possible, is not expedient, since the resulting equation is hard to solve. The method is as follows. Plot the curves 1) in character with sufficient accuracy to determine which of them is steeper (i.e. has the numerically larger slope) at their point of inter- section. Let Ci: F(x,y)=0 be the one that is less steep. or or y =/(«) X = (l>{y), y C>\ y^ \ ■^l \ Hi y ^ \ X a -2 ' "l Fig. 57 the other. Then, making the best guess we can to start with, X = Xi, compute 2/1 from the equation of Cx : and substitute this value in the equation of G^, thus getting the second approximation: Proceeding with a^ in the same manner, we obtain first y^, then X3, and so on. 182 CALCULUS The successive steps of the process are shown geometrically by the broken lines of the figures. The success of the method depends on the ease with which y can be determined when x is given in the case of Ci, while for Q^ X must be easily attainable from y. If the curves hap- pened to have slopes numerically equal but opposite in sign, the process would converge slowly or not at all. But in this case the arithmetic mean of x^ and x^ will obviously give a good approximation. The method has the advantage that each computation is independent of its predecessor. An error, therefore, while it may delay the computation, will not vitiate the result. Example. A beam 1 ft. thick is to be inserted in a panel 10 X 15 ft. as shown in the figure. How long must the beam be made? We have : ' sin 4> + l cos (^ = 15, 15 [ cos ^ + 1 sin <^ = 10. Hence cos" ^ — sin" <^ = 10 cos <^ — 15 sin <^. Fig. 68 Now an expression of the form a cos ^ — 6 sin <^ can always be written as Va^ + V^( "■ cos ^ - ^ sin <^ ) = Va" + 6" cos (<t> + a), where cos a = - In the present case, then, cos 2<^=V325cos (<^ + a), , 10 . 15 where cos a = — , sin « = — • V325 V325 Thus a is an angle of the first quadrant and tan « = I, a = 56° 16'. APPLICATIONS 183 Our problem may be formulated, then, as follows : To find the abscissa of the point of intersection of the curves : y = cos 2 <^, «/ = V326 cos (<^ + a). We know from the figure a good approximation to start with, namely : tan ^ = I, ^= 33° 44'. For this value of <^ the slopes are given by the equations : * i^ . ^ = _ 2 sin 2<^ = - 2 sin 67° 28' = - 1.8, IT d^ i^ . ^ = - V325 sin (<^ + «) = - V325 = - 18. Hence we have : Ci : y — cos 2 <f> ; On-, w = V326 cos (d> + a) or d) = cos^i — ^::rz — a- V325 BegiiiniQg with the approximation ^1 = 33° 44', we compute 2/1 = cos 67° 28' = .3832. Passing now to the curve Cj, we compute its <^ when its .3832 = V326 cos (<^2 + «), <^2 = 32° 31'. We now repeat the process, beginning with <j}2 = 32° 31' and fiiid : 2/2 = cos 65° 02' = .4221, .4221 = V325 cos (.^3 + a), <^3 = 32° 23'. A further repetition gives <^4 = 32° 22', and this is the value of the root we set out to determine. * Since the degree is here taken as the unit of angle, the formulas of differentiation involve the factor ir/lSO ; cf . Chap. V, § 2. 184 CALCULUS EXERCISES 1. Solve the same problem for a beam 2 ft. thick. 2. A cord 1 ft. long has one end fastened at a point 2 ft. above a rough table, and the other end is tied to a rod 2 ft. long. How far can the rod be displaced from the vertical through — and still remain in equilibrium when released ? The equations on which the solution depends are : J 2 cot d + -= cot <j>, I 2 cos 5 + cos <^ = 2. If the coefficient of friction /a = ^, find the value of ^. 3. A heavy ring can slide on a smooth vertical rod. To the ring is fastened a weightless cord of length 2 a, carrying an equal ring knotted at its middle point and having its further end made fast at a distance a from the rod. Find the position of equilibrium of the system. 4. Solve Example 2, § 4, by the method of successive ap- proximations. 7. Arrangement of the Xumerical Work in Tabular Form. In the foregoing paragraphs we have laid the chief stress on setting forth the great ideas which underlie these powerful methods of numerical computation. There are, however, cer- tain details of technique which are important, not only for ease in keeping in view the results obtained, but also for accuracy, since they reduce the numerical work to a system. We will illustrate what we mean by an example. Example. Let it be required to find all the values of x be- tween 0° and 360° which satisfy the equation sin a; = logio (1 — cos x). APPLICATIONS 185 A free-hand graph of each of the functions 1) y = auix, y = logjo (1 — cos x) shows that there is one root between 0° and 180° and a second between 180° and 360°. But these roots cannot be located with any great accuracy iu this manner. It is nec- essary to do exact table work, and to keep the successive results in such form that they are convenient for later reference. To this end such a table as the following is useful.* with the trial value x = 150°. Fig. 60 Begin X 160° cosx 1 — COS a; logio (1 — cos x) - .8660 1.8660 .2709 sinx .5000 Since the ordinate of the sine curve is larger than that of the logarithmic curve, it is clear from the figure that x is too small. Try X = 160°. Before proceeding further let us ask ourselves whether the above scheme is the simplest for the example in hand. Por the special value x = 160° we know cos x without reference to the tables, and hence one entry of the tables was sufficient. But when x = 160°, it will be necessary to enter the tables first for cos x, a second time for logio (1 — cos x), and still a third time for sin x. * Paper ruled in small squares is convenient for these tables, the in- dividual digits being written in separate squares. 186 CALCULUS 1 — cos a; = 2 sin^-, 2' logio (1 — cos x) = login ain^ - + logio 2 = 2 log sin I +.3010. Hence it is possible to get along with, only two entries of tlie tables if we make use of tlie following scheme. X 160° 164° 163° 3' \^ 80° 82° 81° 32' logio sin ^ X 1.9934 T.9958 1.9952 2 logio sin \ X 1.9868 1.9916 1.9904 + .3010 .2878 .2926 .2914 sin (180 - x) .3420 .2756 .2916 The ordinate of the sine curve is still in excess, but only slightly so. Try x ~ 164°. It is seen that the curves have now crossed. Moreover, the two approximations for x — namely, 160° and 164° — are so near together that we can with ad-vantage apply the method of interpolation of § 5. We ■'^ '■ /(o!) = logio (1 — cos a;) ; ^ (a;) = sin x, x^ = 160°, c/,(a:!i) = .3420, <^(k.^) = .2766, 7i = Ai Xi = 164°, f{x^) = .2878, /(a^) = .2926, » .0542 , i = 4°; Ai = .0542 ; A, = .0170; A1 + A2 .0712 : 3.06. Thus the correction is seen to be 3.05°, or 3° 3', and the new approximation is : a; =163° 3'. APPLICATIONS 187 For this value of x the values of the two functions, /(as) and <^ (x), differ by a quantity -which is comparable with the error of the tables, and the problem is solved. EXERCISES 1. Determine the other root in the above problem. 3. Solve the equation : cot X = login (1 + sin a;), < a; < 90°. 3. Find the positive root of the equation e~' = 0? — X. Suggestion. Tabulate x, a? (from a table of cubes), a;' — x, and e""'. 8. Algebraic Equations. By an algebraic equation is meant an equation of the form 1) aoa;" + aiX^-^ + - + a„ = 0, ao¥= 0, where n denotes a positive integer. If the coefBcients Oq, aj, •■■ are numerical, the roots can be approximated to by the method of interpolation or by" Newton's method. In either case it becomes necessary to compute the value of the polynomial f(x) = OqW + aia;»-i + •■ •+ a„ for several values of x, the later ones of which will be at least three- or four-place numbers. There are labor-saving devices for performing these computations, to which we now turn. Numerical Computation of Polynomials. Let a cubic poly- nomial, for example, be given : f(x) = ax^ + bx^ + cx-\-d, and let it be required to compute f{x) for the value x = m. Write down the following scheme : _a am -\- b am^ + bm-\-c f(m) am, am^ + bm am? -\- hm? + cm 188 CALCULUS the explanation of wMeh is as follows. Begin with the first coeflB.cient, a, and multiply it by m to get the expression am which stands below the line. To this expression add the second coefficient, 6, to get the second expression above the line, am + b. Next, multiply this expression by m to get the expression which stands below it, and continue the process. The last entry above the line will be the required value, /(m) = aw? + bm? + <mi + d. Example. Let f{x) = 7a? -&X-- + Zx -1, and let it be required to compute the value of /(«) for x = .8. Here, the scheme is as follows : 7 -.4 2.68 -4.856 5.6 -.32 2.144 and hence /(.8) = - 4.866. It will be observed that the process requires only additions (or subtractions) and multiplications. The former can be per- formed mentally. The latter are executed most simply by one of the machines now in general use with computers. These instruments, combined with the method of this para- graph, have rendered Horner's method for solving numerical algebraic equations obsolete. EXERCISE Compute the value of 5.1a^ - 3.42a;'- -H 1.432a; -|- .8543 for X = .1876. In the problems which arise in physics, however, it is not a question of computing all the roots of a numerical equation, about which nothing is known beyond the coefficients. Usu- ally, the equation is a cubic or biquadratic, and only one root is required. Moreover, from the nature of the problem, a close APPLICATIONS 189 guess at the value of this root can be made at the outset. Then the methods set forth in this paragraph and in §§ 2, 3 lead quickly to the desired result. EXERCISES Solve the following equations, being given that there is one root, and only one, between 0° and 90° : 1. 4 cos3 (9 - 3 cos ^ = .6283, 0° < 6 < 90°. 2. sin' e - .75 sin $ = .1278, 0° < (9 < 90° 3. Find the root of the equation !B* + 2.6SC3 - 5.2a;2 - 10.4a; + 5.0 = which lies between and 1. 4. Find the root of the equation 3x* - 12a? + 12a;2 -4 = which lies between 2 and 3. 9. Continuation. Cubics and Biquadratics. Aside from the special problem of numerical computation, the simpler alge- braic equations present an intrinsic interest which should not be ignored. Transformations, a) Let the cubic equation 1) /(») = aa? + bx'^ + cx + d = 0, a^O, be given, and let x be replaced by y, where 2) y = x — h, x = y -\-h. Then f(x) = a(y + hy + b(y + hf J^c{y + h) + d = ^{y) = af + (3ah + b)y''+-, where the later coefficients are easily written down. 190 CALCULUS If y = )8 is a root of the equation 3) <^(y) = o, then x = p + h •will he a root of equation 1). For, it is always true that /{"') = Hy) when X and y are connected hy the relation 2). Here, h is any number we please. In particidar, h can always he so chosen that the coefficient of the second term of 3) will drop out. It is sufficient to set 4) 3ah + b = 0, or A = -A. Obviously, the same method can be used to transform an algebraic equation of any degree into a new equation whose second term is lacking. EXERCISES Transform the following equations into equations in which the second term is lacking. 1. a.-3 + a;'- — a; + 1 = 0. 2. So? — ix'^ +2 = 0. 3. a;* + a^ — a;2 + l = 0. 4. 5a^ —ia^ + x^ +x — SO=zO. 5. 3x^—7ai' + x^ — x — l = 0. 6. a;6 + a^ + iB2^a,.^l_0. 6) Let the equation 5) f(^x)=x>+pa? + qx + r=0 be given, and let x be replaced by y, where 6) y = ^, x=ky. Then f(x) = kY + k-^ + kqy + r. Denoting this last polynomial by <^ (y), we have fip) = 'f>{y) for all values of x and y which are connected by the relation 6). APPLICATIONS 191 It is clear that, if ^ = j8 is a root of the equation then x = kp will be a root of 5). The factor k is arbitrary, and we can always determine it so that, on dividing equation 7) through by fc : the coefficient of j/' will be numerically equal to unity (provided that p^O): i) ^=1 or k = \/p, if p>0; ii) =2 = — 1 or k = ^/—p, if p <0. k^ In this way, equation 5) can be reduced to one of the two forms a) y* + y^ + Ay + B = 0; /8) y*-y^ + Ay + B = 0. If, in particular, p = and q ^0, 5) can be reduced to the form y) y' + y + B = 0. The method can be applied to any algebraic equation whose second term is lacking : x" + CiX"-^ + CsK"-' 4- - + c„ = 0. EXERCISES 1, Replace the equation 7a!^ - 175a;2 + 16a; + 10 = by an equation of the type j8), and state precisely the relation of the roots of the second equation to those of the first. 192 Calculus 2. Show that, if in the equation Ooa" + aiX°~^ + — + a„ = 0, where Oo =?^ and a„ =^ 0, the transformation 1 2/ = - X is made, the roots of the new equation, a^" + «„-i3r"' H 1- oo = are the reciprocals of the roots of the given equation, 3. If on transforming equation 1) by 2), where h is deter- mined by 4), the constant term in the resulting equation 3), <^(y) = 0, does not vanish, the further transformation 8) ' y = -, or x = - + h, z z will carry 1) into an equation ia which the linear term is lack- ^S- Az^ + B^ + D=:0. A^O, D=^Q. The theorem holds in full generality for an algebraic equa- tion of any higher degree. State it accurately. 4. Replace the equation a^ — 4:a^ — 6af + 16x-4: = by an equation of the type Ay* + Bf+Of' + D = 0. Graphical Treatment. We have already seen that the cubic a^+px + q = can be solved graphically by cuttiag the standard graph y = ai> by the straight liae, •' ° y = -px-q. Since the general cubic can be reduced by the transformation 2) to a cubic of this type, we may consider the general problem of the graphical solution of a cubic as solved. APPLICATIONS 193 To obtain a similar solution for tlie general biquadratic, 9) aoil^ + ba^ + ex"- + dx + e = 0, a 4=0, begin by reducing it to one of the three forms : i) y* + y^ + Ay + B = 0; ii) y*-y2 + Ay + B = 0; iii) y' +Ay + B = 0. An equation of type i) : a^ + x'-+ Ax + B = 0, can be solved graphically by cutting the standard curve y = !e^ by the parabola y =— x' — Ax — B. A similar procedure leads to a solution in the case of each of the other two types, ii) and iii). The Method of Curve Plotting. Let the coefficients a, e in equation 9) be different from 0. By means of Ex. 3, p. 192, the equation can be reduced to one of the following type : J[a;« + B3? + Cx^ + E = (i. In order to discuss the number and location of the roots of this equation, it is sufficient to plot the curve y = Aa^ + Ba?+ Cx^+E. Since all the maxima, minima, and points of inflection of this curve can be determined by means, at most, of quadratic equa- tions, the problem is readily solved in any given numerical case. EXERCISES Determine the number of real roots of each of the following equations, and locate them approximately. 1. 3a!* + 8a;3 -90a:2+ 100 = 0. 2. Sa;* + 8a;' -90a;5 + 600 = 0, 194 CALCULUS 3. 3a!* + 8 a? -90aii2 + 1600 = 0. 4. Show that the eqiiation has rlo real roots. How many real roots has each of the following equations ? 5. a;^— 5x — 1 = 0. 6. a;' + 7a;— 1=0. 7. a? — 4a; + l=0. 8. a;' -3a; — 2 = 0. 9. a;'-x + 3 = 0. 10. 43;=- 15a;2 + 12a;+ 1 = 0. 11. 3a;< + 4a;' + 6a!2-l = 0. 12. 3a;< - 4a?' + 12a;= + 7 = 0. 13. How many positive roots has the equation 6a;* + 8a;3 - 12a;2 - 24a; - 1 = ? 14. Has the equation 3a;»_8a;«+12ar' + l = any real roots ? 16. By means of the graph of the function y = a?+px + q show that the equation a? +px + q = has (a) 1 real root when ^ + — > ; ^7 4 (6) 3 real roots when ^ + 2. < o ; ^ ^ 27 4 ' (c) 2 real roots when ^ + i- = 0, {p and q not both } fd) 1 real root when ^ + 21 = 0, {p = g = 0} ^ ' 27 4 In case (c) it is customary to count one of the roots twice ; in case (d), to count the root three times. APPLICATIONS 195 16. Extend the criterion of Ex. 15 to the case of the general °^^^° aa:? + bofi + cx + d = 0. 10. Curve Plotting. We will close this chapter by consider- ing the application Of the principles set forth in the earlier paragraph-on curve plotting (Chap. Ill, § 6) to some interest- ing curves of a more complex nature. 1) Example 1. To plot the curve 1 1 y- l''"a!-|-l The curve is obviously not symmetric in either axis ; but the test for symmetry in the origin is fulfilled, since on replac- ing a; by — K and yhj—y the new equation, 1 . 1 -y = X — 1 — X + 1 is equivalent to the original equation, 1). Incidentally we observe that the curve passes through the origin. In consequence of the symmetry just noted it will be suffi.cient to plot the curve for positive values of x and then rotate the figure about the origin through 180°. To each positive value of X but one there corresponds one value of y. When X approaches 1 as its limit from above (i.e. always remaining greater than 1), y becomes positively infinite. Hence the line x=l is an asymptote for one branch of the curve. FlQ. 61 196 CALCULUS When X approaches 1 from below, y becomes negatively infinite, and hence this same line, x=l, is an asymptote for a second branch of the curve. Tor all other positive values of x, y is continuous. The slope of the curve is given by the equation 2) ^ = -(. ' dx V' {x-iy (a; + 1)2/ and is seen to be negative for all values of x for which y is continuous. Thus, ia particular, the curve is seen to have no maxima or minima, or in fact any points at which the tangent is horizontal. The second derivative is given by the formula ' dx'' \{x - Vf (x + 1)V When a; > 1, the right-hand side of this equation is always positive, and so the curve is concave upward in this interval. Moreover, it is evident from 1) that, when a; = + oo, y ap- proaches from above, and so the positive axis of x is also an asymptote. In the interval < a; < 1, the second derivative is surely sometimes negative, for this is obviously the case when x is only slightly less than 1. Is dhj/dx^ always negative in this iaterval? If not, it must pass through the value 0; for a continuous function cannot change from a positive to a negative value without taking on the intermediate value 0.* Let us set, then, the right-hand side of equation 3) equal to and solve: 2f— L_ + _^_Vo. * How must the graph of a, continuous function look, which is' some- times positive and sometimes negative ? It must cross the axis of ah- scissas, must it not ? At the point or points where It crosses, the function has the value 0. APPLICATIONS 197 This equation is equivalent to the following : 1 ^ 1 (x-iy (x + iy' Extracting the cube root of each side of this equation, we have : 1 1_ ■ x—1 x+1 Clearing of fractions we find : x + l=-(x-l), or 2a! = 0. Hence a; = is the only value of a! for which (Py/dx'^ can vanish, and we see at once that the right-hand side of 3) does vanish for a! = 0. We have thus proven that the contiauous function 3) is no- where iu the interval < a! <.l, and since it is negative in part of this interval, it is negative throughout. Hence the curve is concave downward throughout the interval. It is now easy to complete the graph. The curve has one point of inflection, — namely, the origin, — and the slope there is, by 2), equal to — 2. EXERCISES Plot the following curves : '■ y-,U- 2. 3a! ^ 3-|-a:2 '■ ^-a!-2 + a! + 2- 4. y=-+ \ X x—1 '■ y-x + x + 1 6. 1 y x^-1 7. y=^- x^ 8. V- ^ (i-xy 1 10. V- ^ ^- ^-a!'" (x + iy 198 CALCULUS 11. y = x-\ 12. y = X X X A. ^ 13. y = —^ + 3? — 2x. 14. y = - .&x — a?. 1 — a; d +» 15.,=^-^. 16. , = 1— A_. a;— la; + l x x—1 Example 2. To plot the curve 4) y^ = x^ + a?. We observe first of all that the curve is symmetric in the axis of a;. It is sufficient, therefore, to plot the e^rve for posi- tive values of y, and then fold this part of the curve over on the axis of x. The curve goes through the origin. Unlike the examples hitherto considered, this curve does not permit ah arbitrary choice of x. It is only when the right- hand side is positive or zero, i.e. when a!2 + a? ^ 0, or a?(l + x)>0 or x>—l, that there will be a corresponding value of y and thus a point with the given abscissa. Obviously, the curve cuts the axis of x at the origin and at the point x = — l. We have, then, essentially two problems : i) to plot the curve for a; > ; ii) to plot the curve for — 1 < a; < 0. i) When a; > 0, the positive value of y Is given by the equation 6) y = xVT+x. Hence dy^ 2 + 3a; _ dx 2Vl-|-a; APPLICATIONS 199 For positive values of x the right-hand side of this equation is always positive, and hence there are no horizontal tangents in the interval under consideration ; the slope of this part of the curve is always positive. In particular, the slope at the origin is unity : dx = 1. 7) The second derivative has the value d^y _ 4 + 3a; <*'«' 4(1 + x)^ The right-hand side of this equation is always positive in this interval, and thus it appears that the curve is concave upward for all posi- ° tive values of x. Fig. 62 ii) When — 1 < a; < 0, the positive value of y is no longer given by the formula 6), since x is now negative.* In the present case, 8) y=~ a; VI + x, and consequently 9) dy _ 2-1- 3a; dx 2V1 -t- X io-> oPy_ 4-1- 3a; i^^"' 4(1 + x)i The first derivative will vanish if, and only if, 2-|-3a; = 0, or x = — ^. * The student must have clearly hi mind the definition of the function expressed hj the y sign, which was laid down in Chap. I, § 1. This func- tion is the positive square root of the radicand ; It can never take on a negative value. 200 CALCULUS It is, therefore, important to determine the corresponding point on the curve and draw the tangent there : 2/|x=-!=-(-|)V5rri=^=.38. Two other important poiots for the present curve are the origin and the point a; = — 1, y = 0. At these points the slope has the following values : -1 .1. dx — 1; dy dx Fto. 63 Draw the corresponding tangents. From the expression 10) for the second derivative it is clear that, when — 1 < a; < 0, the right-hand side of this equation is always negative, and hence the curve is concave down- ward throughout the whole in- terval in question. We can now draw in the curve in this interval, Kg. 63. The curve is now complete above the axis of x. It remains, therefore, merely to fold this part over on that axis. The entire curve is shown in Fig. 64. EXERCISES Fio. 64 Plot the following curves : 1. y^ = x^-a?. 2. y^ = x- Ix"- -|- 7?. 3. y'i={x — a)\Ax-\-B) Suggestion : Write the second factor in the form Jx + B = A(x — b), where b = —, and make two cases : i) A>0; ii) A<0. Discuss the omitted case, ^ = 0. APPLICATIONS 201 4. 2/2 = a;2 - »*. 5. yi — y^^ jgi. Example 3. To plot the curve 11) y'' = x{x-l){x-2). The curve lies wholly in the regions ^ a) ^ 1 and 2 ^ sc. It is symmetric in the axis of x, and hence it is sufficient to plot it for positive values of y. The function y =:-yJx(x — l)(x — 2) is continuous in the interval ^ a; ^ 1. It starts with the value when x = 0, increases, and finally decreases to when a;=l. When X, starting with the " value 2, increases, y, starting with the value 0, increases, always remaining positive, and increasing without limit as x be- comes infinite. So much from considerations of continuity. A piore specific discussion of the character of the curve can be given by means of the derivatives of the function. The slope is given by the formula Fia. 65 12) or 13) 2y^=Zx^-&x + 2 dx dy ^ Zx^—Qx + 2 dx ~ 2^x{x-l){x-2) The slope is infinite when a; = or 1 : = co. dy dx dy dx At these points, the tangent is vertical. 202 CALCULUS The slope is when 3a;2-6a!+2 = 0. The roots of this equation are x=l+—, x = l-—- V3 V3 The first of these values does not correspond to any point on the curve. The second, x = .42, yields a horizontal tangent, the ordinate being " \3V3 3V3 Plot this point and draw the tangent. From the above dis- cussion on the basis of continuity it is obvious that this point must'lbe a maximum, and we see that there are no other maxima or minima. But it is not clear that the curve has no points of inflection in this interval. iG. bb . rpg treat this question, compute the sec- ond derivative. This might be done by means of formula 13) ; but it is simpler to use 12) : 2y^ + 2^=6x-6, y^ = 3x-3-^. "dx^ dx^ Substitute here the value of dy/dx from 13) and reduce : ■' '^dx"' 4 a; (a; -1) (a; -2) And now we seem to be in difficulty. How are we going to tell when dh//dx^ is positive, when negative ? First of all, y is positive, and so the sign of dhjjd^ will be the same as that of the right-hand side of the equation. Secondly, in the interval in question, < a; < 1, the denomi- nator is positive. APPLICATIONS 203 All turns, then, on whether the numerator, i.e. the funotion 15) w = 3!B*-12a!8 + 12a;2-4, is positive or negative. To answer this question, plot the graph of the function 15). The slope of the graph is given by the equation 16) ~ = 12a^ - 36a;! + 24a! = 12 a; (a; - 1) (a; - 2). dx In the interval in question, the right-hand side of this last equation is always positive. Hence u increases with x through- out the interval ^ a; ^ 1, and consequently attains its great- est value at the end-point, x = 1. Here, Mi»=l = -1- We see, therefore, that u is negative throughout the whole interval in question, and consequently the graph of 1) is Concave downward in this interval. The reasoning by which we determined whether u is pos- itive or negative is an excellent illustration of the practical application of the methods of curve plotting which we have learned. It is in no wise a question of the precise values of u which correspond to x. The question is merely : Is u posi- tive, or is it negative? Without the labor of a single com- putation involving table work we have answered this ques- tion with the greatest ease. Such questions as these arise again and again in physics, and the aid which the calculus is able to render here is most important. One further point. It may seem to have been a fluke that we were able to factor the polynomial in 16) and thus simplify so materially the further discussion. And yet, in the problems which arise in practice,' — the problems with a pedigree, — just such simplifications as this present themselves with great frequency. FlQ. 67 204 CALCULUS To complete the graph, it remains to consider the interval 2 ^ a; < oo. Since . the tangent to the curve is vertical at the point where the curve meets the axis of x. It is clear, then, that the curve must be concave downward for a while, and so dPy/da? < for values of X slightly greater than 2. This is verified from 14), since 17) m|^=-4. On the other hand, when x is large, u is positive and (Py/da^ is positive. Hence the curve is concave upward. There must be, therefore, a point of inflection in the interval, and there may be several. From 14) we see that the second derivative will vanish when and only when 3^ _ jgj^ + i2a,2 -4 = 0. The problem is, then, to determine the number of roots of this equation which are greater than 2, and to compute them. Again, it is a question of the graph of 15). When « > 2, we see from 16) that > 0. du dx Hence u steadily increases with X. Now, from 17), u starts with a negative value, and u is positive and large when x is large. Hence u vanishes for just one value of x which is greater than 2. Since m|j_3 = 23, this root is seen to lie between 2 and 3. It can be determined to any required degree of accuracy by the foregoing methods of Fig. 68 APPLICATIONS 205 this chapter, which find herewith a practical application. To two places of decimals it is 2.47. EXERCISES Plot the following curves ; 1. y = a^ — X. 2. y = x — a?. 3. 2/2 = ajs Ij. X. 4.' 2/"! = 1 — ar*. 5.^2/2=(a!2 — l)(a;2-4). 6. y"^ = {1 - x'^){x''- — i). 2 = - « «2 = i ^ OI?-X 9. 7/' '^ . y -1-x w V'- ="' •' 1 + x^' 13. V'- '""' ^ x-1 15. y-^ = a? — A:X'' + 5x. 17. 2/ = sin a;— sin 2a!. 19. y == cos a — cos 2a!. 8. 2/2 = 10. y (a;2 - l)(a!2 - 4) X 1+x 12- t=^^ 14. 2/' = T^- 1 + a; 16. 2/ = sin a; + sin 2 a:. 18. y = cos X + cos 2a!. 20. y = x-\- sin a!, ^ a! ^ tt. CHAPTER VIII THE INVERSE TRIGONOMETRIC FUNCTIONS 1. Inverse Functions. Let (1) y^m be a given function of x, and let us solve this equation for x as a function of y : (2) '^ = 't>(y)- Then <j}(y) is called the inverse function, or the inverse of the function /(k). Thus if /(«) = a^, we have y = af'. Hence x ^ Vy, and <^(2/) is here the function -y^ When the given function is tabulated, the table also serves as a tabulation of the inverse function. It is necessary merely to enter it from the opposite direction. Thus, if we have a table of cubes, we can use it to find cube roots by simply re- versing the roles of the two columns. In the same way, the graph of the fxmction (1) serves as the graph of the function (2), provided in the latter case we take y as the independent variable, and x as the dependent variable, or function. The graph of the inverse function, plotted with x as the in- dependent variable, can be obtained from the former graph as follows. Make the transformation of the plane which is de- fined by the equations : (3) . ^ = ^'1 or ''=y''\ 206 THE INVERSE TRIGONOMETRIC FUNCTIONS 207 Fig. It is easy to interpret this transformation. Any poipt, whose coordinates are (x, y), is carried over into a point {x', y') situ- ated as follows : Draw a line L tlirough the origin bisecting the angle between the positive axes of coordinates. Drop a perpendicular from {x, y) on L and produce it to an equal distance on the other side of L. The poiat thus determined is the point (a;', y'). The proof of this statement is immediately evident from the figure. Thus it appears that the transformation (3) can be generated by rotating the plane about L through 180°. The transformation is also spoken of as a reflection in L, since if a plane mirror were set at right angles to the plane of (x, y) and so that the line L would lie iu the surface of the mirror, the image of any figure, as seen in the mirror, would be the transformed figure. Monotonic Functions. A function, f(x), is said to be mono- tonic if it is single-valued and if, as x increases, /(x) always increases, or else always decreases. We shall be concerned only with functions which are, in general, continuous. It is obvious that the inverse of a monotonic function is also monotonic. A given function, 2/ =/(»'), can in general be considered as made up of a number of pieces, each of which is monotonic in a certain interval.* Thus the function (4) y = sfi * There are functions which do not have this property ; but they do not play an important rdle in the elements of the Calculus. 208 CALCULUS can be taken as made up of two pieces, corresponding respec- tively to those portions of the graph ■which lie in the first and the second quadrants, the corresponding intervals for x being liere -oo<a;<0, 0<a;<oo. Each of the pieces, of which /(cc) is made up, has a monotonic inverse, and thus the function ^(a;) inverse to f(x) is repre- sented by a number of monotonic functions. In the example just cited, the inverse function is multiple- valued : (5) y=±-Vx. But one of the two pieces into which the original function was divided yields the single-valued function (6) y=V5, the so-called principal value of the multiple-valued function (5); the other, y = _V^, the remainder of (6). The derivative of a monotonic function cannot change sign ; but it can vanish or become infinite at special points. Thus y = Va^ — a;2, ^ a; ^ a, is a decreasing monotonic function. Its derivative is, in gen- eral, negative ; but when a; = 0, it vanishes, and- when x=a, it becomes infinite. Differentiation of an Inverse Function. The function <f>{x) inverse to a given function f(x) can be differentiated as follows. By definition, the two equations (7) y=^{x) and x=f(y) are equivalent ; they are two forms of one and the same relar tion between the variables x and y. Their graphs are identical. Take the difEerential of each side of the second equation : dx = d/{y) = DJiy).dy. THE INVERSE TRIGONOMETRIC FUNCTIONS 209 Hence (8) dx DJ{y) To complete the formula, express the right-hand side of (8) in terms of x by means of (7). 2. The Inverse Trigonometric Functions. The inverse trigo- nometric functions are ■ chiefly important because of their application in the Integral Calculus. They are defined as follows. (a) The Function sin~i x. The inverse of the function (1) • y = sin X is obtained as explained in § 1 by solving this equation for x as a function of y, and is written : (!') X = sin~i y, read " the anti-sine of y." * In order to obtain the graph of the function (2) y-=sm~^x ■we have, then, merely to reflect the graph of (1) in the bisector of the angle made by the positive coordinate axes. We are thus led to a multiple-valued func- tion, since the line x = x'(— 1 ^ a;' ^ 1) cuts the graph in more than one point, — in fact, in an infinite number of points. For most purposes of the Calculus, how- ever, it is allowable and advisable to pick * The usual notation^on the Continent for sin-i a;, tan-i x, etc., is arc sin x, arc tan x, etc. It is clumsy, and Is followed for a purely academic reason ; namely, that sin-' x might he misunderstood as meaning the minus first power of sin x. It is seldom that one has occasion to write the recipro- cal of sin X in terms of a negative exponent. When qne wishes to do so, all ambiguity can he avoided by writing (sin x)-K 210 CALCULUS out just one value of the function (2), most simply the value that lies between y = — t/2 and y = + ■rr/2, and to understand by sin~'a! the single-valued function thus obtained. This de- termination is called the 2>rincipal value of the multiple-valued function sin"' a;. Its graph is the portion of the curve in Fig. 70 that is marked by a heavy line. This shall be our convention, then, in the future unless the contrary is explic- itly stated, and thus (3) y = sin"' x is equivalent to the relations : (30 a;=siny, -2^2/<|- In particular, sin-iO = 0, sin-il=^, sin-i(- 1) = -|. The student should now prepare a second plate, showing the graphs of the three functions sin"' x, cos~' as, tan"' w. Place the first in the upper left-hand corner of the sheet ; the second, in the upper right-hand corner ; and the third on the lower half-sheet. All of these curves can be ruled from the templets. Use a fine lead-pencil ; then mark in the principal value of the function in a clean, firm red line. Also mark, in each fig- ure, all the principal points, as is done in Fig. 70 of the text. Differentiation of sin"' x. In order to differentiate the func- tion . _, y = sm ' X, make the equivalent equation, a; = sin 3/ the point of departure. Then dx = d sin y = cos y dy. Hence ' ^ ^. dx cosy THE INVERSE TRIGONOMETRIC FUNCTIONS 211 Tlie right-hand side of this eqiiation. cau be expressed in terms of x as follows. Since sin2 y + C032 y = l and since sin y = x, we have , cos' y = 1 — x% cos y = ± Vl — x''. We have agreed, however, to understand by sin~ia! the principal value of this function. Hence y is subject to the restriction : — ir/2 ^y^ ■jr/2, and consequently cos y is posi- tive (or zero). We must, therefore, take- the upper sign before the radical,* the final result thus being : (4) d . _, 1 — sin 1 X = — dx Vl- (4') d sin~i X = dx Vl-a? (b) The Function cos~ia;. The treatment here is precisely similar. The definition is as follows : (5) y = cos~i X if x = cos y, (read : " anti-cosine x "). The graph of the function cos~ia; is as shown in Fig. 71. Like sin^ia;, this function is also infinitely multiple-valued. A single- valued branch is selected by imposing the further condition O^y^ir. This determination is known as the principal value of cos~i x : (6) y- O^y^TT. * Geometrically the slope of the portion of the graph in question is always positive, and so we must use the positive square root of 1 — x''. 212 CALCULUS It will be understood lieiiceforth. that the principal value is meant unless the contrary is explicitly stated. In preparing the graph of this function, mark the principal value as a firm red line. To differentiate the function cos~ia!, use the implicit form of equation (5) : X = cos y. Hence dx = d cos y = — sin y dy 3>iid. dj _ dx 1 siny For the principal value, sin^ is positive. and hence (V d — cos~ia; = dx 1 , Vl-a;2 or (7') d cos~i X = dx vr The principal values of the functions sin~* x and cos~i x are connected by the identical relation : (8) sin~i9!+ cos~ia; = ^- By means of this relation, the differentiation of cos"* a could have been performed immediately. (c) The Function tan"* x. Here, the definition is as follows : (9) 2/ = tan~ia; if a; = tany, (read : " anti-tangent x "). The principal value is defined as that determination which lies between — Tr/2 and Tr/2 : (10) 2/ = tan-ia!, _|<2,<|. In preparing the graph of this function, mark the principal value as a firm red line. THE INVERSE TRIGONOMETRIC FUNCTIONS 213 y ^ . 2 /^ 03 ^ T^ 3 Fia. 72 To differentiate tan"** use the implicit form (9). Hence da; = d tan y = sec^ y dy, dy^ 1 da: sec^ y Since sec^ y = 1-\- tan^ y and tan y = x, it follows that (11) or (12) d tan-la; = -^. ^ ' 1 4- a;'' (d) 2%e Function cot-i a;. Here, the definition is : (13) y = cot-i a; if x = cot y, (read : " anti-cotangent x "). The principal value is chosen as that one which lies between and TT ; (14) y >= cot-' X, < y < TT. — tan-i X = , da; 1 + a;2' 214 CALCULUS The differentiation can be performed as in the case of the function tan"ia;, but still more simply by means of the identi- cal relation connecting the principal values of tan"'* and cot"ia;: (15) tan-* X + cot"* « = f. Hence (16) lco1r>a; = ^, ^ ' dx 1 + 3^' or (17) ci!cot-ia; = ^. It is well for the student to make a graph of this function, also, drawing in the principal value, as usual, in red. The following identity holds for positive values of x, when the principal values of the functions are used : (18) tan-i- = cot-i x, 0<x. X For negative values of x it reads : (180 tan-ii = cot-i x-tt, x<0. X Remarks. The other inverse trigonometric functions, sec-'o!, cacr^x, can be treated in a similar manner. They are, how- ever, without importance in practice. Their principal values cannot be defined by means of a single continuous curve. The graph necessarily consists of more than one piece ; it is most natural to take it as consisting of two pieces. Corresponding to the Addition Theorem for each of the trigonometric functions, there are functional relations for the inverse trigonometric functions. Thus, for tan-'a! : (19) tan-i u + tan-i v = tan-i '" + ^ • ^ ' 1 — WW These relations, however, are not always true when the principal value of each of the functions is taken, and for this THE INVERSE TRIGONOMETRIC FUNCTIONS 215 reason it is usually better not to employ them. If, however, ia a particular case, u and v are each numerically less than unity, the principal values can be used throughout in (19). 3. Shop Work. The student will now add to his list of Special Formulas the four new formulas of this chapter. The list of formulas of dififerentiation is now complete. It reads as follows. Special Formulas of Differentiation 1. dc = 0. 2. d a" = na;"~' da;. 3. d sin x = cos x dx. 4. d cos x = — sin x dx. 5. dtan X = sec' x dx. 6. d cot x = — esc' X dx. 7. 8. 9. 10. 11. 12. 13. dloga; X de" = e^da;. da" = a' log a da;. d sin~^ X dx Vl-a;2 d cos~^ X dx VI- ^ d tan"' SB da; 1+ a;2 d cot-la;: da; l + a;2 It is important that the student gain facility in the use of the new results. 216 CALCULUS Exa/mple 1. Differentiate the function u = cos~i - , ■ a > 0. Let y =s a Then u = cos"' y, du = d cos~i y — ^y Hence — - , dx dy = — a dx dy a dx VI — f r. 7«Y Va2 - X' and, finally, dC08-^ = - ''«' « Va2 — a!2 In abbreviated form. «) 4 , a cos""^ - = - Example 2. Differentiate the function Va2 — ! « = tan-i2«' + l. 3 Here, d 20! + 1 3 idx 3dx 1 + r2x + v K 3 , f 10 + 4x4-43!!' / 9 ^5+2x + 2!ifi du or dx 5 + 2x + 2a^ THE INVERSE TRIGONOMETRIC FUNCTIONS 217 The student should notice that the method used in the text for deriving the fundamental formulas of differentiation is not to he repeated in the applications. It is these formulas them- selves that should be used. Thus, to solve Ex. 1 by writiag X cos u = - a and then differentiating would be logically irreproachable, but bad technique. EXERCISES Differentiate each of the following functions. 1. M = smi — — iH-. --^11 a 2. M = tan~i — a 3. M = cot~'-- a 4. M = sin~i(w sin x). -1 1 — « 5. M = cos 1 — - — • , , 2 dx V3-|-2a; — a;2 . ,2a! — 1 „ i.-,x+a 6. M = sm-i=^^^^ — =• 7. ?t = coti--f — V2 * 8. M = cot-1 = • 9. M = tan-'-- X 10. u = tan ' 1 — X- dx V^ — x- du 1 dx a^ + X"' du = ''~a^ dx + x^ du ncosa; dx vr — in? sitf X du 1 ;=:tan-V a;-— ; 11. u^tssr^ix^ — — -„ ■ 3a;' ■ _,x — a 12. M = sm 1 X du 2 dx l-|-a;2 du 3 dx~ 1 + a;' 13 . u = : C0S"1 — X X + a du 2 ^ dx Vl — a;* V2 218 15. « = C0S-1-. 2 16. t=sin-ii. 17. «=cos-i-+Y. n 18. u= X sin^i a;. 1 CALCULUS ^ = -2siii2t dt - = 3cos3<. (if ds . . — = — nsiii?i(<- -y)- 19. „-tan-ix X du 1 20. u = - _ ^ . ^^^^ <^ Vr;=^(siii-ix)2 21. M = acos-i^^l^. 22. M = tan-i ^~" . a rc + a 23. M = cot-i^±^. 24 «=.sm-i55^±^. 03! — a bx + a 25. M=V^^^a^-acos-i^. *f = X?lEZ, « > 0. a; dx X 26. w = sm-i? + ^^^H^ ** = ^^EZ,a>0 ax dx x^ du 2 27. M = tan-i/'2tan-^ 28. M = tan-i(3 tan 6). da; 5 — 3 cos x du 3 de 5 — 4cos2i9 4. Continnation. Numerical Computation. By means of the Tables the numerical value of any of the functions of this chapter can be determined when a specific numerical value has been chosen for the independent variable. It is, however, an important aid to ease and security in such computations to be able, in advance, to make sure of the early significant figures and the location of the decimal point. There are two impor- tant geometrical methods for achieving this end. One js the representation of the trigonometric functions by suitable lines connected with the unit circle ; the other consists in the graphs introduced above, in § 2. THE INVERSE TRIGONOMETRIC FUNCTIONS 219 iFirBt of all, however, it should be pointed out that there are two distinct problems. One is to find all values of x which satisfy such equations as (a) sin a; = .2318 ; (6) cos a; = - .4322 ; (c) tan a; = - 1.4861. The other is to find the principal value of an inverse trigono- metric function ; for example, sin-1.2318; cos-i (- .4322) ; tan-i(- 1.4861) The methods of treating these problems are identical- First Geometric Method. Equations (a), (6), (c) can be solved graphically by the aid of the unit circle representation with an error corresponding to a degree or two, the results being expressed in radians if the problem comes from the Calculus. For example, consider equation (6). The student should provide himself with an accurately drawn circle of his own construction, executed on the accurate centimeter-millimeter paper commercially procurable ; the radius of the circle berug 10 cm. and its center at a principal intersection of the rulings. To solve equation (6), he will lay a straight-edge on his plate, parallel to the secondary (or y-) axis and at a distance of 4 cm., 3^ mm. to the left of that axis. Marking the two points of intersection of the straight-edge with the circle by fine pencil lines easUy erased, he now measures one of the acute angles involved by means of his protractor and thus determines the two solutions of (6) lying between 0° and 360° correct to minutes or thereabouts. By aid of the Tables the values can at once be converted into radian measure. Arithmetic Solutions. From the figure before him the stu- dent now sees clearly a right triangle, one leg of which is known. The determination of the angle he needs is merely a problem in the solution of a right triangle by the tables, and 220 CALCULUS he proceeds to carry this work through to the degree of accu- racy which the tables permit. Equations (a) and (c) are treated in a similar manner. The point of this method is that the student is trained to visualize a figure, and not to try to remember a table that looks like sin^ + + — — . For, such tables vanish ia a short time, and when the student needs his trigonometry ia later work, he is helpless. In terms of the inverse functions, this first problem consists in finding all the values of the multiple-valued function cos~i x for the value of the variable, x = — .4322. Second Oeometric Method. This method consists ia readiag off from the graph the two values which lie between and 2 jr, and then adding to these arbitrary positive or negative multi- ples of 2 jr. The graph suggests, moreover, how to determine these values arithmetically by the aid of a table of sines or cosines of angles of the first quadrant. It also suggests a further refinement of the graphical method, of which the student will do well to avail himself, — namely, this. Let him make an accurate graph of the function ° ^ y— sin X on cm.-mm.-paper, taking 10 cm. as the unit and measuring the angle in radians, x ranging from to ir/2. This half-arch supplements the four graphs of the functions sin x, cos x, sin~i x, cos~ia; and serves as a 3-place table for determining their values (with a possible error of two or three units in the third place). To sum up, then, there are two geometric methods ; 1) the unit-circle method ; 2) the graphs of the functions, the latter beiag supplemented by the 10-cm. graph just described. Either of the geometric methods suggests how to use the tables correctly and affords an altogether satisfactory check on the tables. When the accurately drawn graphs are not at hand, free- hand drawings indicate clearly how to use the tables with security and accuracy. THE INVERSE TRIGONOMETRIC FUNCTIONS 221 EXERCISES 1. Determine both in degrees and radians all values of x which satisfy the above equations (a), (6), (c), using each time all of the geometric methods set forth, and also the tables. 2. Mnd the value of each of the following functions. It is understood that the principal value is meant. Use first the method of the graphs. Then determine from the tables. Check by unit-circle and protractor. i) sin-i(-.1643); ii) cos-i (.6417) ; iii) tan-i(- 2.8162). 3. By means of a free-hand drawing of the graph estimate the value of each of the following functions. Remember that a curve recedes from its tangent very slowly near a point of inflection. a) sin-1.113; 6) tan-i(- .214) ; c) cos-i.l72; d) tan-i (- 7.4) ; e) cot"! ( - .162) ; /) cos-i (- .998) ; g) sin-i(-.21); h) sin-i.89; i) tan-i6.2; j) cot-17.3; fc) cos-i(-.138); Z) sin-i (- .138). In what cases is your error large ; in what, small ? 5. Applications. The inverse trigonometric functions afford a convenient means of solving the following problem in Optics. A ray of light is refracted in a prism. Show that its deviation from its original direction is least when the incident ray and the refracted ray make equal angles with the faces of the prism. The study of this problem has a pj^ ^3 vivid interest for the student who has seen the laboratory experiment of admitting a ray of sunlight into a darkened room, allowing it to pass through 222 CALCULUS a prism, thus being refracted, and throwing it finally, dis- persed, on a screen. Let AF be the incident ray; PQ, its path through the prism ; and Q£ the ray -which emerges. Then the deflection of PQ is obviously 6— ^ and the further deflection of Q,B is 6' — <l>' ; so that the total deflection, u, is : (1) u = e-4, + e'-4>' = e+6'-(ct> + <i>'). On the other hand, the sum of the angles of the triangle PDQ is / \ / \ '=(i-*Hi-*')+- Hence (2) ^ + 0'=a. We can, therefore, write (1) in the form : (3) u = e+e'-a. This is the quantity it is desired to make a minimum. and 6' are, however, connected by a relation which can be obtained as follows. We have by the law of refraction (cf. Chap. V, § 7) : /A\ sin 5 (') sin^-'^' sin 6' -: ;= n. sin<^' Letv=l/M. Then (6) sin <^ = V sin fl or <l> = sin~i (v sin 8). Similarly, (6) sin <^' = i/ sine' or ^' = sin-i(i/sine') Substituting these values of ^ and </>' in equation (2) we have the desired relation : (7) sin-i (v sin ff) + sin-i (v sin 0') = a. Our problem now is completely formulated ; it is : To make the function u given by (3) a minimum, when 8 and 0' are con- nected by (7) : u = + 0'-a, sin"i (v sin 0) + sin~i (v sin 0') =a. (8) THE INVERSE TRIGONOMETRIC FUNCTIONS 223 Take 6 as the independent variable. Then ^' M-^ + W and the condition — - = gives -^ = — 1. de ^ d0 Next, take the differential of each side of the second equar tion (8) : d(vsing) d(vsiD.6') ^ Vl — y2 sin2 d Vl — 1/2 sia2 6' or V cos 6dd , V cos ff d6' _ « Hence VI — 1/2 giji2 Q VI _ y2 sui2 01 (■\n\ cos g . cos^^ /'^^''\_o Vl - v2 sin2 6 Vl-i/2sin2 0'Uey But dO'/dO = — 1. Consequently /-. -| ^ cos 5 cos 6' VI — v2 sin2 a Vl - v^ sin2 e' One solution of this equation is 6 = 6', — the solution de- manded by the theorem. But conceivably there might be other solutions, and then it -would not be clear which one of them makes u a miaimum. We can readUy show, however, that equation (11) has no further solutions. Square each side : cos^g ^ cos^y l-v2siii2e l-v-'Sia^e'' Clear of fractions and express each cosine in terms of the sine : (1 - v2 sin2 0')(1 - sin' S)=(l- v"- sitf e)(l - sia^ 6'). Multiply out and suppress equal terms on the two sides : _ sin2 e - v"' sin2 & = - sin' ff — v^ sin' B, if - 1) sin2 e = (v2 - 1) sin2 6'. 224 CALCULUS Hence siii2 e = siii2 fl', sin 6 = sin 6', and consequently the only angles of the first quadrant which can satisfy (11) are equal angles, 6 = 6'. From (5) and (6) it follows that <f> = <^'. Hence, from (2) (^ = ^, and so $= sin-i( n sin^j, M = 2sin"i( Msin- )— a. \ V That M is a minimum, is clearly indicated by the labora- tory experiment. It can be proven analytically as follows. From (9) dW- dfl2 ' Differentiate (10) as it stands ; then, after the differentia- tion, set dd'/dd = — 1 and & = 6'. It is seen at once tha*- ^'<0, hence ^ > 0, and u has a minimum. EXERCISE The bottom of a mural painting 4 ft. high is 12 ft. above the eye of the observer. How far back from the wall should he stand, in order that the angle subtended by the painting be as large as possible ? Suggestion. Take the distance, x, of the observer from the wall as the independent variable, and express the angle of elevation of the bottom and the top of the painting in terms of X. Printed in the United States of America.